Sample Exercises

Hint

Kelvin and Celsius scales are related as:
\(
T_{\mathrm{C}}=T_{\mathrm{K}}-273.15 \ldots \text { (i) }
\)
Celsius and Fahrenheit scales are related as:
\(
T_F=\frac{9}{5} T_c+32 \ldots \text { (ii) }
\)
For neon:
\(
\begin{aligned}
& T_{\mathrm{K}}=24.57 \mathrm{~K} \\
& \therefore T_{\mathrm{C}}=24.57-273.15=-248.58^{\circ} \mathrm{C} \\
& T_F=\frac{9}{5} T_c+32 \\
& \frac{9}{5}(-248.58)+32 \\
& =-415.44^{\circ} \mathrm{F}
\end{aligned}
\)
For carbon dioxide:
\(
\begin{aligned}
& T_{\mathrm{K}}=216.55 \mathrm{~K} \\
& \therefore T_{\mathrm{C}}=216.55-273.15=-56.60^{\circ} \mathrm{C} \\
& T_F=\frac{9}{5}\left(T_c\right)+32 \\
& =\frac{9}{5}(-56.60)+32 \\
& =-69.88^{\circ} \mathrm{C}
\end{aligned}
\)

Hint

Triple point of water on absolute scale A, \(T_1=200 \mathrm{~A}\)
Triple point of water on absolute scale \(\mathrm{B}, T_2=350 \mathrm{~B}\)
Triple point of water on Kelvin scale, \(T_K=273.15 \mathrm{~K}\)
The temperature \(273.15 \mathrm{~K}\) on Kelvin scale is equivalent to \(200 \mathrm{~A}\) on absolute scale A.
\(
\begin{aligned}
& T_1=T_K \\
& 200 \mathrm{~A}=273.15 \mathrm{~K} \\
& \therefore A=\frac{273.15}{200}
\end{aligned}
\)
The temperature \(273.15 \mathrm{~K}\) on Kelvin scale is equivalent to \(350 \mathrm{~B}\) on absolute scale B.
\(
\begin{aligned}
& T_2=T_K \\
& 350 \mathrm{~B}=273.15 \\
& \therefore B=\frac{273.15}{350}
\end{aligned}
\)
\(T_{\mathrm{A}}\) is triple point of water on scale A.
\(T_{\mathrm{B}}\) is triple point of water on scale B.
\(
\begin{aligned}
& \therefore=\frac{273.15}{200} \times T_A=\frac{273.15}{350} \times T_B \\
& T_A=\frac{200}{350} T_B
\end{aligned}
\)
Therefore, the ratio \(T_A: T_B\) is given as 4: 7 .

Hint

Here, \(R_0=101.6 \Omega ; T_0=273.16 \mathrm{~K}\) Case (i) \(R_1=165.5 \Omega ; T_1=600.5 \mathrm{~K}\), Case (ii) \(R_2=123.4, T_2=\) ?
Using the relation \(R=R_0\left[1+\alpha\left(T-T_0\right)\right]\)
Case (i) \(165.5=101.6[1+\alpha(600.5-273.16)]\)
\(
\begin{aligned}
& \alpha=\frac{165.5-101.6}{101.6 \times(600.5-273.16)}=\frac{63.9}{101.6 \times 327 \times 37} \\
& \text { Case II } 123.4=101.6\left[1+\alpha\left(T_2-273.16\right)\right] \\
& \text { or } 123.4=101.6\left[1+\frac{63.9}{101.6 \times 327.34}\left(T_2-273.16\right)\right] \\
& =101.6+\frac{63.9}{327.37}\left(T_2-273.16\right) \\
& \text { or } T_2=\frac{(123.4-101.6) \times 327.34}{63.9}+273.16=111.67+273.16 \\
& =384.83 \mathrm{~K}
\end{aligned}
\)

Hint

Let \(T_F\) be the temperature on Fahrenheit scale and \(T_K\) be the temperature on absolute scale. Both the temperatures can be related as:
\(
\frac{T_F-32}{180}=\frac{T_K-273.15}{100} \dots(i)
\)
Let \(T_{F 1}\) be the temperature on Fahrenheit scale and \(T_{K 1}\) be the temperature on absolute scale. Both the temperatures can be related as:
\(
\frac{T_{F 1}-32}{180}=\frac{T_{K 1}-273.15}{100} \dots(ii)
\)
It is given that:
\(
T_{K 1}-T_K=1 K
\)
Subtracting equation (i) from equation (ii), we get:
\(
\begin{aligned}
& \frac{T_{F 1}-T_F}{180}=\frac{T_{K 1}-T_K}{100}=\frac{1}{100} \\
& T_{F 1}-T_F=\frac{1 \times 180}{100}=\frac{9}{5}
\end{aligned}
\)
Triple point of water \(=273.16 \mathrm{~K}\)
\(\therefore\) Triple point of water on absolute scale \(=273.16 \times \frac{9}{5}=491.69\)

Hint

Triple point of water, \(T=273.16 \mathrm{~K}\).
At this temperature, pressure in thermometer A, \(P_A=\mathbf{1 . 2 5 0} \times \mathbf{1 0}{ }^5 \mathrm{~Pa}\)
Let \(T_1\) be the normal melting point of sulphur.
At this temperature, pressure in thermometer \(\mathrm{A}, P_B=1.797 \times 10^5 \mathrm{~Pa}\)
According to Charles’ law, we have the relation:
\(
\begin{aligned}
& \frac{P_A}{T}=\frac{P_1}{T_1} \\
& \therefore T_1=\frac{P_1 T}{P_A}=\frac{1.797 \times 10^5 \times 273.16}{1.250 \times 10^5} \\
& =392.69 \mathrm{~K}
\end{aligned}
\)
Therefore, the absolute temperature of the normal melting point of sulphur as read by thermometer A is \(392.69 \mathrm{~K}\).
At triple point \(273.16 \mathrm{~K}\), the pressure in thermometer \(\mathrm{B}, P_B=0.200 \times 10^5 \mathrm{~Pa}\)
At temperature \(T_1\), the pressure in thermometer \(\mathrm{B}, P_2=0.287 \times 10^5 \mathrm{~Pa}\)
According to Charles’ law, we can write the relation:
\(
\begin{aligned}
& \frac{P_B}{T}=\frac{P_1}{T_1} \\
& \frac{0.200 \times 10^5}{273.16}=\frac{0.287 \times 10^5}{T_1} \\
& \therefore T_1=\frac{0.287 \times 10^5}{0.200 \times 10^5} \times 273.16=391.98 \mathrm{~K}
\end{aligned}
\)

Therefore, the absolute temperature of the normal melting point of sulphur as read by
thermometer B is 391.98 K.

Hint

Length of the steel tape at temperature \(T=27^{\circ} \mathrm{C}, l=1 \mathrm{~m}=100 \mathrm{~cm}\)
At temperature \(T_1=45^{\circ} \mathrm{C}\), the length of the steel rod, \(l_1=63 \mathrm{~cm}\)
Coefficient of linear expansion of steel, \(\alpha=1.20 \times 10^{-5} \mathrm{~K}^{-1}\)
Let \(l_2\) be the actual length of the steel rod and \(l^{\prime}\) be the length of the steel tape at \(45^{\circ} \mathrm{C}\).
\(
\begin{aligned}
& l^{\prime}=l+a l\left(T_1-T\right) \\
& \therefore l^{\prime}=100+1.20 \times 10^{-5} \times 100(45-27) \\
& =100.0216 \mathrm{~cm}
\end{aligned}
\)
Hence, the actual length of the steel rod measured by the steel tape at \(45^{\circ} \mathrm{C}\) can be calculated as:
\(
l_2=\frac{100.0216}{100} \times 63=63.0136 \mathrm{~cm}
\)
Therefore, the actual length of the rod at \(45.0^{\circ} \mathrm{C}\) is \(63.0136 \mathrm{~cm}\). Its length at \(27.0^{\circ} \mathrm{C}\) is \(63.0 \mathrm{~cm}\).

Hint

The given temperature, \(T=27^{\circ} \mathrm{C}\) can be written in Kelvin as: \(27+273=300 \mathrm{~K}\)
The outer diameter of the steel shaft at \(T, d_1=8.70 \mathrm{~cm}\)
Diameter of the central hole in the wheel at \(T, d_2=8.69 \mathrm{~cm}\)
Coefficient of linear expansion of steel, \(\boldsymbol{\alpha}_{\text {steel }}=1.20 \times 10^{-5} \mathrm{~K}^{-1}\)
After the shaft is cooled using ‘dry ice’, its temperature becomes \(T_1\).
The wheel will slip on the shaft, if the change in diameter, \(\Delta d=8.69-8.70\) \(=-0.01 \mathrm{~cm}\)
Temperature \(T_1\), can be calculated from the relation:
\(
\begin{aligned}
& \Delta d=d_1 \boldsymbol{\alpha}_{\text {steel }}\left(T_1-T\right) \\
& 0.01=8.70 \times 1.20 \times 10^{-5}\left(T_1-300\right) \\
& \left(T_1-300\right)=95.78 \\
& \therefore T_1=204.21 \mathrm{~K} \\
& =204.21-273.16 \\
& =-68.95^{\circ} \mathrm{C}
\end{aligned}
\)
Therefore, the wheel will slip on the shaft when the temperature of the shaft is \(-69^{\circ} \mathrm{C}\).

Hint

Initial temperature, \(T_1=27.0^{\circ} \mathrm{C}\)
Diameter of the hole at \(T_1, d_1=4.24 \mathrm{~cm}\)
Final temperature, \(T_2=227^{\circ} \mathrm{C}\)
Diameter of the hole at \(T_2=d_2\)
Co-efficient of linear expansion of copper, \(\alpha_{C u}=1.70 \times 10^{-5} \mathrm{~K}^{-1}\)
For co-efficient of superficial expansion \(\beta\), and change in temperature \(\Delta T\), we have the relation:
\(\frac{\text { Change in area }(\triangle A)}{\text { Original area }(A)}=\beta_{\Delta T}\)
\(\frac{\left(\pi \frac{d_2^2}{4}-\pi \frac{d_1^2}{4}\right)}{\left(\pi \frac{d_1^2}{4}\right)}=\frac{\Delta A}{A}\)
\(\therefore \frac{\Delta A}{A}=\frac{d_2^2-d_1^2}{d_1^2}\)
But \(\beta=2 a\)
\(
\begin{aligned}
& \therefore \frac{d_2^2-d_1^2}{d_1^2}=2 a \Delta T \\
& \frac{d_2^2}{d_1^2}-1=2 a\left(T_2-T_1\right) \\
& \frac{d_2^2}{d(4.24)^2}=2 \times 1.7 \times 10^{-5}(227-27)+1 \\
& d_2^2=17.98 \times 1.0068=18.1 \\
& \therefore d_2=4.2544 \mathrm{~cm}
\end{aligned}
\)
Change in diameter \(=d_2-d_1=4.2544-4.24=0.0144 \mathrm{~cm}\)
Hence, the diameter increases by \(1.44 \times 10^{-2} \mathrm{~cm}\).

Hint

Initial temperature, \(T_1=27^{\circ} \mathrm{C}\)
Length of the brass wire at \(T_1, l=1.8 \mathrm{~m}\)
Final temperature, \(T_2=-39^{\circ} \mathrm{C}\)
Diameter of the wire, \(d=2.0 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}\)
Tension developed in the wire \(=F\)
Coefficient of linear expansion of brass, \(\alpha=2.0 \times 10^{-5} \mathrm{~K}^{-1}\)
Young’s modulus of brass, \(Y=0.91 \times 10^{11} \mathrm{~Pa}\)
Young’s modulus is given by the relation:
\(
\begin{aligned}
& Y=\frac{\text { Stress }}{\text { Strain }}=\frac{\frac{F}{A}}{\frac{\Delta L}{L}} \\
& \Delta L=\frac{F \times L}{A \times Y} \dots(i)
\end{aligned}
\)
Where,
\(F=\) Tension developed in the wire
\(A=\) Area of cross-section of the wire.
\(\Delta L=\) Change in the length, given by the relation:
\(\Delta L=\alpha L\left(T_2-T_1\right) \ldots\) (ii)
Equating equations (i) and (ii), we get:
\(
\begin{aligned}
& a L\left(T_2-T_1\right)=\frac{F L}{\pi\left(\frac{d}{2}\right)^2 \times Y} \\
& F=a\left(T_2-T_1\right) \pi\left(\frac{d}{2}\right)^2 \\
& F=2 \times 10^{-5} \times(-39-27) \times 3.14 \times 0.91 \times 10^{11} \times\left(\frac{2 \times 10^{-3}}{2}\right)^2 \\
& =-3.8 \times 10^2 N \\
& 3.8 \times 10^2 \mathrm{~N}
\end{aligned}
\)
(The negative sign indicates that the tension is directed inward.)
Hence, the tension developed in the wire is \(3.8 \times 10^2 \mathrm{~N}\).

Hint

Initial temperature, \(T_1=40^{\circ} \mathrm{C}\)
Final temperature, \(T_2=250^{\circ} \mathrm{C}\)
Change in temperature, \(\Delta T=T_2-T_1=210^{\circ} \mathrm{C}\)
Length of the brass rod at \(T_1, l_1=50 \mathrm{~cm}\)
Diameter of the brass rod at \(T_1, d_1=3.0 \mathrm{~mm}\)
Length of the steel rod at \(T_2, l_2=50 \mathrm{~cm}\)
Diameter of the steel rod at \(T_2, d_2=3.0 \mathrm{~mm}\)
Coefficient of linear expansion of brass, \(\alpha_1=\mathbf{2 . 0} \times 10^{-5} \mathrm{~K}^{-1}\)
Coefficient of linear expansion of steel, \(\alpha_2=1.2 \times 10^{-5} \mathrm{~K}^{-1}\)
For the expansion in the brass rod, we have:
\(
\begin{aligned}
& \frac{\text { Change in area }(\triangle A)}{\text { Original area }(A)}=a_1 \Delta T \\
& \therefore \Delta l_1=50 \times\left(2.1 \times 10^{-5}\right) \times 210 \\
& =0.2205 \mathrm{~cm}
\end{aligned}
\)
For the expansion in the steel rod, we have:
\(
\begin{aligned}
& \frac{\text { Change in area }(\Delta A)}{\text { Original area }(A)}=a_2 \Delta T \\
& \therefore \Delta l_2=50 \times\left(2.1 \times 10^{-5}\right) \times 210 \\
& =0.126 \mathrm{~cm}
\end{aligned}
\)
Total change in the lengths of brass and steel,
\(
\begin{aligned}
& \Delta l=\Delta l_1+\Delta l_2 \\
& =0.2205+0.126
\end{aligned}
\)
\(=0.346 \mathrm{~cm}\)
Total change in the length of the combined rod \(=0.346 \mathrm{~cm}\)
Since the rod expands freely from both ends, no thermal stress is developed at the junction.

Hint

Coefficient of volume expansion of glycerin, \(\alpha_V=49 \times 10^{-5} \mathrm{~K}^{-1}\)
Rise in temperature, \(\Delta T=30^{\circ} \mathrm{C}\)
Fractional change in its volume \(=\frac{\Delta V}{V}\)
This change is related with the change in temperature as:
\(
\begin{aligned}
& \frac{\Delta V}{V}=a_v \Delta T \\
& V_{T_2}-V_{T_1}=V_{T_1} a_v \Delta T \\
& \frac{m}{\rho_{T_2}}-\frac{m}{\rho_{T_1}}=\frac{m}{\rho_{T_1}} a_v \Delta T
\end{aligned}
\)
Where,
\(
\begin{aligned}
& m=\text { Mass of glycerine } \\
& \rho_{T_1}=\text { Initial density at } T_1 \\
& \rho_{T_2}=\text { Final density at } T_2 \\
& \frac{\rho_{T_1}-\rho_{T_2}}{\rho_{T_2}} a_v \Delta T
\end{aligned}
\)
Where,
\(\frac{\rho_{T_1}-\rho_{T_2}}{\rho_{T_2}}=\) Fractional change in density
\(\therefore\) Fractional change in the density of glycerin \(=49 \times 10^{-5} \times 30=1.47 \times 10^{-2}\)

Hint

Power of the drilling machine, \(P=10 \mathrm{~kW}=10 \times 10^3 \mathrm{~W}\)
Mass of the aluminum block, \(m=8.0 \mathrm{~kg}=8 \times 10^3 \mathrm{~g}\)
Time for which the machine is used, \(t=2.5 \mathrm{~min}=2.5 \times 60=150 \mathrm{~s}\)
Specific heat of aluminium, \(c=0.91 \mathrm{~J} \mathrm{~g}^{-1} \mathbf{K}^{-1}\)
Rise in the temperature of the block after drilling \(=\delta T\)
Total energy of the drilling machine \(=P_t\)
\(
\begin{aligned}
& =10 \times 10^3 \times 150 \\
& =1.5 \times 10^6 \mathrm{~J}
\end{aligned}
\)
It is given that only \(50 \%\) of the power is useful.
Useful energy, \(\Delta Q=\frac{50}{100} \times 1.5 \times 10^6=7.5 \times 10^5 \mathrm{~J}\)
But \(\Delta Q=m c \Delta T\)
\(
\begin{aligned}
& \therefore \Delta T=\frac{\Delta Q}{m c} \\
& =\frac{7.5 \times 10^5}{8 \times 10^3 \times 0.91} \\
& =103^{\circ} \mathrm{C}
\end{aligned}
\)
Therefore, in \(2.5\) minutes of drilling, the rise in the temperature of the block is \(103^{\circ} \mathrm{C}\).

Hint

Mass of the copper block, \(m=2.5 \mathrm{~kg}=2500 \mathrm{~g}\)
Rise in the temperature of the copper block, \(\Delta \theta=500^{\circ} \mathrm{C}\)
Specific heat of copper, \(C=0.39 \mathrm{~J} \mathrm{~g}^{-1} \mathrm{C}^{-1}\)
Heat of fusion of water, \(L=335 \mathrm{~J} \mathrm{~g}^{-1}\)
The maximum heat the copper block can lose, \(Q=m C \Delta \theta\)
\(
\begin{aligned}
= & 2500 \times 0.39 \times 500 \\
& =487500 \mathrm{~J}
\end{aligned}
\)
Let \(m_1 \mathrm{~g}\) be the amount of ice that melts when the copper block is placed on the ice block.
The heat gained by the melted ice, \(Q=m_1 L\)
\(
\therefore m_1=\frac{Q}{L}=\frac{487500}{335}=1455.22 \mathrm{~g}
\)
Hence, the maximum amount of ice that can melt is \(1.45 \mathrm{~kg}\).

Hint

Mass of the metal, \(m=0.20 \mathrm{~kg}=200 \mathrm{~g}\)
Initial temperature of the metal, \(T_1=150^{\circ} \mathrm{C}\)
Final temperature of the metal, \(T_2=40^{\circ} \mathrm{C}\)
Calorimeter has water equivalent of mass, \(m^{\prime}=0.025 \mathrm{~kg}=25 \mathrm{~g}\)
Volume of water, \(V=150 \mathrm{~cm}^3\)
Mass \((M)\) of water at temperature \(T=27^{\circ} \mathrm{C}\) :
\(
150 \times 1=150 \mathrm{~g}
\)
Fall in the temperature of the metal:
\(
\Delta T=T_1-T_2=150-40=110^{\circ} \mathrm{C}
\)
Specific heat of water, \(C_w=4.186 \mathrm{~J} / \mathrm{g} /{ }^{\circ} \mathrm{K}\)
Specific heat of the metal \(=C\)
Heat lost by the metal, \(\theta=m C \Delta T \ldots\). (i)
Rise in the temperature of the water and calorimeter system:
\(
\Delta T^{\prime}=40-27=13^{\circ} \mathrm{C}
\)
Heat gained by the water and calorimeter system:
\(
\begin{aligned}
& \Delta \theta^{\prime}=m_1 C_w \Delta T^{\prime} \\
& =\left(M+m^{\prime}\right) C_w \Delta T^{\prime} \ldots \text { (ii) }
\end{aligned}
\)
Heat lost by the metal \(=\) Heat gained by the water and colorimeter system
\(
\begin{aligned}
& m C \Delta T=\left(M+m^{\prime}\right) C_w \Delta T^{\prime} \\
& 200 \times C \times 110=(150+25) \times 4.186 \times 13 \\
& \therefore C=\frac{175 \times 4.186 \times 13}{110 \times 200}=0.43 J_{g^{-1}} K^{-1}
\end{aligned}
\)
If some heat is lost to the surroundings, then the value of \(C\) will be smaller than the actual value.

Hint

The initial temperature of the body of the child, \(T_1=101^{\circ} \mathrm{F}\)
The final temperature of the body of the child, \(T_2=98^{\circ} \mathrm{F}\)
Change in temperature, \(\Delta T=\left[\left(101-98 \times \frac{5}{9}\right)\right]{ }^{\circ} \mathrm{C}\)
Time taken to reduce the temperature, \(t=20 \mathrm{~min}\)
Mass of the child, \(m=30 \mathrm{~kg}=30 \times 10^3 \mathrm{~g}\)
Specific heat of the human body \(=\) Specific heat of water \(=c\)
\(=1000 \mathrm{cal} / \mathrm{kg} /{ }^{\circ} \mathrm{C}\)
Latent heat of evaporation of water, \(L=580 \mathrm{cal} \mathrm{g} \mathrm{g}^{-1}\)
The heat lost by the child is given as: \(\triangle \theta=m c \triangle T\)
\(
\begin{aligned}
& =30 \times 1000 \times(101-98) \times \frac{5}{9} \\
& =50000 \mathrm{cal}
\end{aligned}
\)
Let \(m_1\) be the mass of the water evaporated from the child’s body in \(20 \mathrm{~min}\).
Loss of heat through water is given by:
\(
\begin{aligned}
& \triangle \theta=m_1 L \\
& \therefore m_1=\frac{\triangle \theta}{L} \\
& =\frac{50000}{580}=86.2 g \\
& \therefore \text { Average rate of extra evaporation caused by the drug }=\frac{m_1}{t} \\
& =\frac{86.2}{200}=4.3 \mathrm{~g} / \mathrm{min}
\end{aligned}
\)

Hint

Side of the given cubical ice box, \(s=30 \mathrm{~cm}=0.3 \mathrm{~m}\)
Thickness of the ice box, \(l=5.0 \mathrm{~cm}=0.05 \mathrm{~m}\)
Mass of ice kept in the ice box, \(m=4 \mathrm{~kg}\)
Time gap, \(t=6 \mathrm{~h}=6 \times 60 \times 60 \mathrm{~s}\)
Outside temperature, \(T=45^{\circ} \mathrm{C}\)
Coefficient of thermal conductivity of thermacole, \(K=0.01 \mathrm{~J} \mathrm{~s}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-}\)
Heat of fusion of water, \(L=335 \times 10^3 \mathrm{~J} \mathrm{~kg}^{-1}\)
Let \(m^{\prime}\) be the total amount of ice that melts in \(6 \mathrm{~h}\).
The amount of heat lost by the food: \(\theta=\frac{K A(T-0) t}{l}\)
Where
\(
\begin{aligned}
& A=\text { Surface area of the box }=6 s^2=6 \times(0.3)^2=0.54 \mathrm{~m}^3 \\
& \theta=\frac{0.01 \times 0.54 \times(45) \times 6 \times 60 \times 60}{0.05}=104976 \mathrm{~J}
\end{aligned}
\)
But \(\theta=m \prime L\)
\(
\begin{aligned}
& \therefore m \prime=\frac{\theta}{L} \\
& =\frac{104976}{335 \times 10^3}=0.313 \mathrm{~kg}
\end{aligned}
\)
Mass of ice left \(=4-0.313=3.687 \mathrm{~kg}\)
Hence, the amount of ice remaining after \(6 \mathrm{~h}\) is \(3.687 \mathrm{~kg}\).

Hint

Here \(K=109 J s^{-1} m^{-1} K^{-1}\)
\(
\begin{aligned}
& A=0.15 \mathrm{~m}^2 \\
& d=1.0 \mathrm{~cm}=10^{-2} \mathrm{~m} \\
& T_2=100^{\circ} \mathrm{C}
\end{aligned}
\)
Let \(T_1=\) temperature of the part of the boiler in contact with the stove.
if \(Q\) be the amount of heat flowing per second throught the base of the boiler, then
\(
\begin{aligned}
& Q=\frac{109 \times 0.15 \times\left(T_1-100\right)}{10^{-2}} \dots(i)\\
& =1635\left(T_1-100\right) J_{s^{-1}}
\end{aligned}
\)
Also heat of vaporisation of water \(L=2256 \times 10^3 \mathrm{Jkg}^{-1}\)
Rate of boiling of water in the boiler, \(\mathrm{M}=6.0 \mathrm{kgmin}=\frac{-1}{60}=0.0 \mathrm{~kg}^{-1} \mathrm{~s}\)
\(\therefore\) Heat received by water per second, \(\mathrm{Q}=\mathrm{ML}\)
\(\Rightarrow Q=0.1 \times 2256 \times 10^3 J_s^{-1} \ldots\) (ii)
:. From equation \(\mathrm{i}\) and (ii) we get
\(
1635\left(T_1-100\right)=2256 \times 10^2
\)
or \(T_1-100=\frac{2256 \times 10^2}{1653}=138\)
\(
T_1=138+100=238^{\circ} \mathrm{C}
\)

Hint

According to Newton’s law of cooling, the rate of cooling is proportional to the difference in temperature.
Here Average of \(80^{\circ} \mathrm{C}\) and \(50^{\circ} \mathrm{C}=65^{\circ} \mathrm{C}\)
Temperature of surroundings \(=20^{\circ} \mathrm{C}\)
\(\therefore\) Difference \(=65-20=45^{\circ} \mathrm{C}\)
Under these condition. the body cools \(30^{\circ} \mathrm{C}\) in time 5 minutes
\(\therefore \frac{\text { Change in temp }}{\text { Time }}=K \triangle T\) or \(\frac{30}{5}=K \times 45^{\circ} \dots(i)\)
The average of \(60^{\circ} \mathrm{C}\) and \(30^{\circ}\) is \(45^{\circ} \mathrm{C}\) which is \(25^{\circ} \mathrm{C}(45-20)\) above the room temperature anf the bodycppls by \(30^{\circ} C(60-30)\) in time \(t\) (say)
\(
\therefore \frac{30}{t}=K \times 25 \dots(ii)
\)
Where \(\mathrm{K}\) is same for this situation as for the original.
Dividing equation i by ii we get
\(
=\frac{30 / 5}{30 / \mathrm{t}}=\frac{K \times 45}{K \times 25}
\)
or \(\frac{t}{5}=\frac{9}{5}\)
\(\Rightarrow t=9 \mathrm{~min}\)

Hint

\(
\begin{aligned}
t & =\frac{p-p_0}{p_{100}-p_0} \times 100^{\circ} \mathrm{C} \\
& =\frac{77.8-73}{100^{\circ} 3-73} \times 100^{\circ} \mathrm{C}=17^{\circ} \mathrm{C} .
\end{aligned}
\)

Hint

The temperature in kelvin is defined as
\(
T=\frac{p}{p_{t r}} \times 273.16 \mathrm{~K}
\)
Thus,
\(
\begin{aligned}
373 \cdot 15 & =\frac{1.50 \times 10^4 \mathrm{~Pa}}{p_{t r}} \times 273.16 \\
p_{t r} & =1.50 \times 10^4 \mathrm{~Pa} \times \frac{273 \cdot 16}{373 \cdot 15} \\
& =1.10 \times 10^4 \mathrm{~Pa}
\end{aligned}
\)

Hint

The room temperature on the scale measured by the thermometer is
\(
t=\frac{p_t-p_0}{p_{100}-p_0} \times 100^{\circ} \mathrm{C} .
\)
Thus,
\(
20^{\circ} \mathrm{C}=\frac{p_t-73.00 \mathrm{~cm} \text { of } \mathrm{Hg}}{100 \mathrm{~cm} \text { of } \mathrm{Hg}-73.00 \mathrm{~cm} \text { of } \mathrm{Hg}} \times 100^{\circ} \mathrm{C}
\)
or, \(\quad p_t=78.4 \mathrm{~cm}\) of mercury.

Hint

For an ideal gas at constant volume,
\(
\begin{aligned}
& \frac{T_1}{T_2}=\frac{p_1}{p_2} \\
& T_2=\frac{p_2}{p_1} T_1 .
\end{aligned}
\)
\(
\text { or, } \quad T_2=\frac{p_2}{p_1} T_1 \text {. }
\)
The temperature of melting ice at 1 atm is \(273 \cdot 15 \mathrm{~K}\). Thus, the temperature of the liquid is
\(
T_2=\frac{160}{80} \times 273.15 \mathrm{~K}=546.30 \mathrm{~K} .
\)

Hint

At \(27^{\circ} \mathrm{C}\), the pressure is \(75 \mathrm{~cm}+5 \mathrm{~cm}=80 \mathrm{~cm}\) of mercury. At the liquid temperature, the pressure is \(75 \mathrm{~cm}+45 \mathrm{~cm}=120 \mathrm{~cm}\) of mercury. Using \(T_2=\frac{P_2}{P_1} T_1\), the temperature of the liquid is
\(
\begin{aligned}
T & =\frac{120}{80} \times(27 \cdot 0+273 \cdot 15) \mathrm{K}=450 \cdot 22 \mathrm{~K} . \\
& =177 \cdot 07^{\circ} \mathrm{C} \approx 177^{\circ} \mathrm{C} .
\end{aligned}
\)

Hint

The temperature on the platinum scale is defined as
\(
t=\frac{R_t-R_0}{R_{100}-R_0} \times 100^{\circ} .
\)
The boiling point of sulphur on this scale is
\(
t=\frac{6 \cdot 50-2 \cdot 50}{3 \cdot 50-2 \cdot 50} \times 100^{\circ}=400^{\circ} .
\)

Hint

The resistances of the wire in the thermometer at \(100^{\circ} \mathrm{C}\) and \(50^{\circ} \mathrm{C}\) are
\(
R_{100}=R_0\left[1+\alpha \times 100^{\circ} \mathrm{C}+\beta \times\left(100^{\circ} \mathrm{C}\right)^2\right]
\)
and, \(\quad R_{\text {b } 0}=R_0\left[1+\alpha \times 50^{\circ} \mathrm{C}+\beta \times\left(50^{\circ} \mathrm{C}\right)^2\right]\).
The temperature \(t\) measured on the platinum thermometer is given by
\(
\begin{aligned}
t & =\frac{R_{50}-R_0}{R_{100}-R_0} \times 100^{\circ} \\
& =\frac{\alpha \times 50^{\circ} \mathrm{C}+\beta \times\left(50^{\circ} \mathrm{C}\right)^2}{\alpha \times 100^{\circ} \mathrm{C}+\beta \times\left(100^{\circ} \mathrm{C}\right)^2} \times 100^{\circ} \\
& =50.4^{\circ} \mathrm{C}.
\end{aligned}
\)

Hint

Let \(R_{t_p}\) denote the resistance of the coil at the platinum scale temperature \(t_p\). Then
\(
\begin{aligned}
t_p & =\frac{R_{t_p}-R_0}{R_{100}-R_0} \times 100 \\
\text { or, } \quad R_{t_p} & =\frac{t_p}{100}\left(R_{100}-R_0\right)+R_0 \\
& =\frac{t_p}{100}\left[R_0\left\{1+\alpha \times 100+\beta \times(100)^2\right\}-R_0\right]+R_0 \\
& =\frac{t_p}{100}\left[\alpha \times 100+\beta \times(100)^2\right] R_0+R_0 \\
& =R_0\left[1+\left\{\alpha \times 100+\beta \times(100)^2\right\} \frac{t_p}{100}\right] \\
& =R_0\left[1+\alpha t_p+\beta \times(100) t_p\right] .
\end{aligned}
\)
Only numerical values of \(\alpha\) and \(\beta\) are to be used.

Hint

The length of the iron rod at \(100^{\circ} \mathrm{C}\) is
\(
\begin{aligned}
l_1 & =(50 \mathrm{~cm})\left[1+\left(12 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\right)\left(100^{\circ} \mathrm{C}-20^{\circ} \mathrm{C}\right)\right] \\
& =50.048 \mathrm{~cm} .
\end{aligned}
\)
The length of the aluminium rod at \(100^{\circ} \mathrm{C}\) is
\(
\begin{aligned}
l_2 & =(100 \mathrm{~cm})\left[1+\left(24 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\right)\left(100^{\circ} \mathrm{C}-20^{\circ} \mathrm{C}\right)\right] \\
& =100 \cdot 192 \mathrm{~cm} .
\end{aligned}
\)
The length of the composite system at \(100^{\circ} \mathrm{C}\) is
\(
50 \cdot 048 \mathrm{~cm}+100 \cdot 192 \mathrm{~cm}=150 \cdot 24 \mathrm{~cm} \text {. }
\)
The length of the composite system at \(20^{\circ} \mathrm{C}\) is \(150 \mathrm{~cm}\). So, the average coefficient of linear expansion of the composite rod is
\(
\begin{aligned}
\alpha & =\frac{0.24 \mathrm{~cm}}{150 \mathrm{~cm} \times\left(100^{\circ} \mathrm{C}-20^{\circ} \mathrm{C}\right)} \\
& =20 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1} .
\end{aligned}
\)

Hint

The ring should be heated to increase its diameter from \(15 \cdot 00 \mathrm{~cm}\) to \(15.05 \mathrm{~cm}\).
Using \(\quad l_2=l_1(1+\alpha \Delta \theta)\),
\(
\begin{aligned}
& =\frac{0.05 \mathrm{~cm}}{15.00 \mathrm{~cm} \times 12 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}} \\
& =278^{\circ} \mathrm{C} \\
\text { The temperature } & =20^{\circ} \mathrm{C}+278^{\circ} \mathrm{C}=298^{\circ} \mathrm{C} .
\end{aligned}
\)
\(
\text { The strain developed }=\frac{l_2-l_1}{l_1}=3.33 \times 10^{-3} \text {. }
\)

Hint

The time period at temperature \(\theta\) is
\(
\begin{aligned}
T & =2 \pi \sqrt{l_\theta / g} \\
& =2 \pi \sqrt{l_0(1+\alpha \theta) / g} \\
& =2 \pi \sqrt{l_0 / g}(1+\alpha \theta)^{1 / 2} \\
& \approx T_0\left(1+\frac{1}{2} \alpha \theta\right) .
\end{aligned}
\)
Thus, \(\quad T_{20}=T_0\left[1+\frac{1}{2} \alpha\left(20^{\circ} \mathrm{C}\right)\right]\)
and, \(\quad T_{40}=T_0\left[1+\frac{1}{2} \alpha\left(40^{\circ} \mathrm{C}\right)\right]\)
or, \(\quad \frac{T_{40}}{T_{20}}=\left[1+\left(20^{\circ} \mathrm{C}\right) \alpha\right]\left[1+\left(10^{\circ} \mathrm{C}\right) \alpha\right]^{-1}\)
\(\approx\left[1+\left(20^{\circ} \mathrm{C}\right) \alpha\right]\left[1-\left(10^{\circ} \mathrm{C}\right) \alpha\right]\)
\(\approx 1+\left(10^{\circ} \mathrm{C}\right) \alpha\)
or, \(\quad \frac{T_{40}-T_{20}}{T_{20}}=\left(10^{\circ} \mathrm{C}\right) \alpha=1.2 \times 10^{-4} \dots(i)\).
This is fractional loss of time. As the temperature increases, the time period also increases. Thus, the clock goes slow. The time lost in 24 hours is, by (i),
\(
\Delta t=\left(24 \text { hours }\left(1^{.} 2 \times 10^{-4}\right)=10^{.} 4 \mathrm{~s}\right. \text {. }\)

Hint

The time period at temperature \(\theta\) is
\(
\begin{aligned}
T & =2 \pi \sqrt{l_\theta / g} \\
& \approx T_0\left(1+\frac{1}{2} \alpha \theta\right)
\end{aligned}
\)
Thus, \(\quad T_{20}=T_0\left[1+\alpha\left(10^{\circ} \mathrm{C}\right)\right]\)
or, \(\quad \frac{\left(T_{20}-T_0\right)}{T_0}=\alpha\left(10^{\circ} \mathrm{C}\right) \dots(i)\)
\(T_{20}\) is the correct time period. The period at \(0^{\circ} \mathrm{C}\) is smaller so that the clock runs fast. Equation (i) gives approximately the fractional gain in time. The time gained in 24 hours is
\(
\begin{array}{rlrl}
\Delta T & =(24 \text { hours })\left[\left(10^{\circ} \mathrm{C}\right) \alpha\right] \\
\text { or, } & 15 \mathrm{~s} & =(24 \text { hours })\left[\left(10^{\circ} \mathrm{C}\right) \alpha\right]
\end{array}
\)
\(
\text { or, } \quad \begin{aligned}
\alpha & =\frac{15 \mathrm{~s}}{(24 \text { hours })\left(10^{\circ} \mathrm{C}\right)} \\
& =1.7 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1} .
\end{aligned}
\)

Hint

Let the volume of the metal piece be \(V_0\) at \(27^{\circ} \mathrm{C}\) and \(V_\theta\) at \(42^{\circ} \mathrm{C}\). The density of the liquid at \(27^{\circ} \mathrm{C}\) is \(\rho_0=1.24 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}\) and the density of the liquid at \(42^{\circ} \mathrm{C}\) is \(\rho_\theta=1.20 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}\).

The weight of the liquid displaced \(=\) apparent loss in the weight of the metal piece when dipped in the liquid. Thus,
\(
V_0 \rho_0=46 \mathrm{~g}-30 \mathrm{~g}=16 \mathrm{~g}
\)
and, \(\quad V_\theta \rho_\theta=46 \mathrm{~g}-30.5 \mathrm{~g}=15 \cdot 5 \mathrm{~g}\).
Thus,
or,
\(
\frac{V_\theta}{V_0}=\frac{\rho_0}{\rho_\theta} \times \frac{15.5}{16}
\)
or, \(1+3 \alpha\left(42^{\circ} \mathrm{C}-27^{\circ} \mathrm{C}\right)=1.00104\)
or, \(\quad \alpha=2.3 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\).

Hint

It is given that the expansion of the sphere is negligible as compared to the expansion of the liquid. At \(0^{\circ} \mathrm{C}\), the density of the liquid is \(\rho_0=1.527 \mathrm{~g} \mathrm{~cm}^{-3}\). At \(35^{\circ} \mathrm{C}\), the density of the liquid equals the density of the sphere. Thus,
\(
\begin{aligned}
\rho_{35} & =\frac{266.5 \mathrm{~g}}{\frac{4}{3} \pi(3.5 \mathrm{~cm})^3} \\
& =1.484 \mathrm{~g} \mathrm{~cm}^{-3}
\end{aligned}
\)
We have
or,
\(
\begin{array}{ll}
\text { We have } & \frac{\rho_\theta}{\rho_0}=\frac{V_0}{V_\theta}=\frac{1}{(1+\gamma \theta)} \\
\text { or, } & \rho_\theta=\frac{\rho_0}{1+\gamma \theta} . \\
\text { Thus, } & \gamma=\frac{\rho_0-\rho_{35}}{\rho_{35}\left(35^{\circ} \mathrm{C}\right)}
\end{array}
\)
\(
\begin{aligned}
& =\frac{(1.527-1.484) \mathrm{g} \mathrm{cm}^{-3}}{\left(1.484 \mathrm{~g} \mathrm{~cm}^{-3}\right)\left(35^{\circ} \mathrm{C}\right)} \\
& =8.28 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1} .
\end{aligned}
\)

Hint

Suppose the length of the iron rod at \(0^{\circ} \mathrm{C}\) is \(l_{i_0}\) and the length of the copper rod at \(0^{\circ} \mathrm{C}\) is \(l_{c 0^{\circ}}\). The lengths at temperature \(\theta\) are
\(
\begin{aligned}
& l_{i \theta}=l_{i 0}\left(1+\alpha_i \theta\right) \dots(i) \\
& l_{c \theta}=l_{c 0}\left(1+\alpha_c \theta\right) \text {. } \dots(ii) \\
& \text { and } \\
& \text { Subtracting, } \\
& l_{i \theta}-l_{c \theta}=\left(l_{i 0}-l_{c 0}\right)+\left(l_{i 0} \alpha_i-l_{c 0} \alpha_c\right) \theta . \dots(iii) \\
& l_{i \theta}-l_{c \theta}=l_{i 0}-l_{c 0}(=10 \mathrm{~cm}) . \\
& \text { Thus, from (iii), } l_{i 0} \alpha_i=l_{c 0} \alpha_c \\
& \text { or, } \\
& \frac{l_{i 0}}{l_{c 0}}=\frac{\alpha_c}{\alpha_i} \\
& \text { or, } \quad \frac{l_{i 0}}{l_{i 0}-l_{c 0}}=\frac{\alpha_c}{\alpha_c-\alpha_i} \\
& =\frac{1.7 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}}{0.6 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}}=\frac{17}{6} . \\
&
\end{aligned}
\)
This shows that \(l_{i 0}-l_{c 0}\) is positive. Its value is \(10 \mathrm{~cm}\) as given in the question.
\(
\begin{aligned}
& \text { Hence, } \quad l_{i 0}=\frac{17}{6} \times\left(l_{i 0}-l_{c 0}\right) \\
& =\frac{17}{6} \times 10 \mathrm{~cm}=28.3 \mathrm{~cm} \\
& \text { and } \\
& l_{c 0}=l_{i 0}-10 \mathrm{~cm}=18 \cdot 3 \mathrm{~cm} . \\
&
\end{aligned}
\)

Hint

Let us assume that the tension is zero at \(100^{\circ} \mathrm{C}\) so that \(l_\theta\) is the natural length of the wire at \(100^{\circ} \mathrm{C}\). As the wire cools down, its natural length decreases to \(l_0\). As the wire is fixed at the clamps, its length remains the same as the length at \(100^{\circ} \mathrm{C}\). Thus, the extension of the wire over its natural length at \(0^{\circ} \mathrm{C}\) is
\(
l_\theta-l_0=l_0(1+\alpha \theta)-l_0=l_0 \alpha \theta \text {. }
\)
The strain developed is \(\frac{l_\theta-l_0}{l_\theta} \approx \frac{l_\theta-l_0}{l_0}=\alpha \theta\).
The stress developed \(=Y \times\) strain \(=Y \alpha \theta\).
The tension in the wire at \(0^{\circ} \mathrm{C}\) is
\(
\begin{aligned}
T & =\text { stress } \times \text { area } \\
& =Y \alpha t \times 0.20 \mathrm{~mm}^2 \\
& =\left(2.0 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}\right) \times\left(1.2 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right) \\
& \times 100^{\circ} \mathrm{C} \times\left(0.20 \times 10^{-6} \mathrm{~m}^{-2}\right) \\
= & 48 \mathrm{~N} .
\end{aligned}
\)
This is equal to the extra force exerted by each clamp.

Hint

The volume of mercury at \(25^{\circ} \mathrm{C}\) is
\(
V_0=100 \mathrm{~cm}^{-3} \text {. }
\)
The coefficient of volume expansion of mercury
\(
\gamma_L=1.8 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}
\)
The coefficient of volume expansion of glass
\(
\begin{aligned}
\gamma_S & =3 \times 1.8 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1} \\
& =5.4 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}
\end{aligned}
\)
Thus, the volume of mercury at \(75^{\circ} \mathrm{C}\) is
\(
V_{L \theta}=V_0\left(1+\gamma_L \Delta \theta\right)
\)
and the volume of the vessel at \(75^{\circ} \mathrm{C}\) is
\(
V_{S \theta}=V_0\left(1+\gamma_S \Delta \theta\right) \text {. }
\)
The volume of mercury overflown
\(
\begin{aligned}
& =V_{L \theta}-V_{S \theta}=V_0\left(\gamma_L-\gamma_S\right) \Delta \theta \dots(i) \\
& =\left(100 \mathrm{~cm}^{-3}\right)\left(1.8 \times 10^{-4}-5.4 \times 10^{-6}\right) /{ }^{\circ} \mathrm{C} \times\left(50^{\circ} \mathrm{C}\right) \\
& =0.87 \mathrm{~cm}^3 .
\end{aligned}
\)
Note that \(\gamma_a=\left(\gamma_L-\gamma_S\right)\) acts as the effective coefficient of expansion of the liquid with respect to the solid. The expansion of mercury ‘as seen from the glass’ can be written as
\(
\begin{array}{rlrl}
V_\theta-V_0 & =V_0 \gamma_a \theta \\
\text { or, } & V_\theta & =V_0\left(1+\gamma_a \theta\right) .
\end{array}
\)
The constant \(\gamma_a\) is called the ‘apparent coefficient of expansion’ of the liquid with respect to the solid.

Hint

The \(75 \mathrm{~cm}\) length of steel at \(0^{\circ} \mathrm{C}\) will become \(l_\theta\) at \(30^{\circ} \mathrm{C}\) where,
\(
l_\theta=(75 \mathrm{~cm})\left[1+\alpha\left(30^{\circ} \mathrm{C}\right)\right] \dots(i).
\)
The length of mercury column at \(30^{\circ} \mathrm{C}\) is \(l_\theta\). Suppose the length of the mercury column, if it were at \(0^{\circ} \mathrm{C}\), is \(l_0\). Then,
\(
l_\theta=l_0\left[1+\frac{1}{3} \gamma\left(30^{\circ} \mathrm{C}\right)\right] \dots(ii).
\)
By (i) and (ii),
\(
l_0\left[1+\frac{1}{3} \gamma\left(30^{\circ} \mathrm{C}\right)\right]=75 \mathrm{~cm}\left[1+\alpha\left(30^{\circ} \mathrm{C}\right)\right]
\)
\(
\text { or, } \quad \begin{aligned}
l_0 & =75 \mathrm{~cm} \frac{\left[1+\alpha\left(30^{\circ} \mathrm{C}\right)\right]}{\left[1+\frac{1}{3} \gamma\left(30^{\circ} \mathrm{C}\right)\right]} \\
& \approx 75 \mathrm{~cm}\left[1+\left(\alpha-\frac{\gamma}{3}\right)\left(30^{\circ} \mathrm{C}\right)\right] \\
& =74.89 \mathrm{~cm} .
\end{aligned}
\)

Hint

The heat released as the water cools down from \(25^{\circ} \mathrm{C}\) to \(0^{\circ} \mathrm{C}\) is
\(
Q=m s \Delta \theta=(0-2 \mathrm{~kg})\left(4200 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}\right)(25 \mathrm{~K})=21000 \mathrm{~J} .
\)
The amount of ice melted by this much heat is given by
\(
m=\frac{Q}{L}=\frac{21000 \mathrm{~J}}{3 \cdot 4 \times 10^5 \mathrm{~J} \mathrm{~kg}^{-1}}=62 \mathrm{~g} .
\)

Hint

Let \(L\) be the specific latent heat of vaporization of water. The mass of the steam condensed is \(1.5 \mathrm{~g}\). Heat lost in condensation of steam is
\(
Q_1=(1.5 \mathrm{~g}) L \text {. }
\)
The condensed water cools from \(100^{\circ} \mathrm{C}\) to \(30^{\circ} \mathrm{C}\). Heat lost in this process is
\(
Q_2=(1.5 \mathrm{~g})\left(1 \mathrm{cal} \mathrm{g}^{-1 \circ} \mathrm{C}^{-1}\right)\left(70^{\circ} \mathrm{C}\right)=105 \mathrm{cal} .
\)
Heat supplied to the calorimeter and to the cold water during the rise in temperature from \(25^{\circ} \mathrm{C}\) to \(30^{\circ} \mathrm{C}\) is
\(
Q_3=(15 \mathrm{~g}+165 \mathrm{~g})\left(1 \mathrm{cal} \mathrm{g}{ }^{-1}{ }^{\circ} \mathrm{C}^{-1}\right)\left(5^{\circ} \mathrm{C}\right)=900 \mathrm{cal} .
\)
If no heat is lost to the surrounding,
\(
\begin{aligned}
\text { (1.5 g) } L+105 \mathrm{cal} & =900 \mathrm{cal} \\
L & =530 \mathrm{cal} \mathrm{g}^{-1} .
\end{aligned}
\)

Hint

Heat required to take the ice from \(-10^{\circ} \mathrm{C}\) to \(0^{\circ} \mathrm{C}\)
\(
=(1 \mathrm{~kg})\left(2100 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}\right)(10 \mathrm{~K})=21000 \mathrm{~J} \text {. }
\)
Heat required to melt the ice at \(0^{\circ} \mathrm{C}\) to water
\(
=(1 \mathrm{~kg})\left(3.36 \times 10^5 \mathrm{~J} \mathrm{~kg}^{-1}\right)=336000 \mathrm{~J} \text {. }
\)
Heat required to take \(1 \mathrm{~kg}\) of water from \(0^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\)
\(
=(1 \mathrm{~kg})\left(4200 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}\right)(100 \mathrm{~K})=420000 \mathrm{~J} \text {. }
\)
Heat required to convert \(1 \mathrm{~kg}\) of water at \(100^{\circ} \mathrm{C}\) into steam
\(
=(1 \mathrm{~kg})\left(2.25 \times 10^6 \mathrm{~J} \mathrm{~kg}^{-1}\right)=2.25 \times 10^6 \mathrm{~J} \text {. }
\)
Total heat required \(=3.03 \times 10^6 \mathrm{~J}\).

Hint

The heat given by the water when it cools down from \(30^{\circ} \mathrm{C}\) to \(0^{\circ} \mathrm{C}\) is
\(
(0.01 \mathrm{~kg})\left(4200 \mathrm{~J} \mathrm{~kg}^{-1}{ }^{\circ} \mathrm{C}^{-1}\right)\left(30^{\circ} \mathrm{C}\right)=1260 \mathrm{~J} .
\)
The heat required to bring the ice to \(0^{\circ} \mathrm{C}\) is
\(
(0.005 \mathrm{~kg})\left(2100 \mathrm{~J} \mathrm{~kg}^{-1}{ }^{\circ} \mathrm{C}^{-1}\right)\left(20^{\circ} \mathrm{C}\right)=210 \mathrm{~J} .
\)
The heat required to melt \(5 \mathrm{~g}\) of ice is
\(
(0.005 \mathrm{~kg})\left(3.36 \times 10^5 \mathrm{~J} \mathrm{~kg}^{-1}\right)=1680 \mathrm{~J} \text {. }
\)
We see that whole of the ice cannot be melted as the required amount of heat is not provided by the water. Also, the heat is enough to bring the ice to \(0^{\circ} \mathrm{C}\). Thus the final temperature of the mixture is \(0^{\circ} \mathrm{C}\) with some of the ice melted.

Hint

Total heat supplied to the system in 4 minutes is \(Q=100 \mathrm{cal} \mathrm{s}^{-1} \times 240 \mathrm{~s}=2.4 \times 10^4 \mathrm{cal}\).

The heat required to take the system from \(-20^{\circ} \mathrm{C}\) to \(0^{\circ} \mathrm{C}\)
\(
\begin{aligned}
& \quad=(100 \mathrm{~g}) \times\left(0-2 \mathrm{cal} \mathrm{g}^{-1}{ }^{\circ} \mathrm{C}^{-1}\right) \times\left(20^{\circ} \mathrm{C}\right)+ \\
& (200 \mathrm{~g}) \times\left(0-5 \mathrm{cal} \mathrm{g}^{-1}{ }^{\circ} \mathrm{C}^{-1}\right) \times\left(20^{\circ} \mathrm{C}\right) \\
& =400 \mathrm{cal}+2000 \mathrm{cal}=2400 \mathrm{cal} .
\end{aligned}
\)
The time taken in this process \(=\frac{2400}{100} \mathrm{~s}=24 \mathrm{~s}\).
The heat required to melt the ice at \(0^{\circ} \mathrm{C}\)
\(
=(200 \mathrm{~g})\left(80 \mathrm{cal} \mathrm{g}^{-1}\right)=16000 \mathrm{cal} .
\)
The time taken in this process \(=\frac{16000}{100} \mathrm{~s}=160 \mathrm{~s}\).
If the final temperature is \(\theta\), the heat required to take the system to the final temperature is
\(
=(100 \mathrm{~g})\left(0.2 \mathrm{cal} \mathrm{g}^{-1}{ }^{\circ} \mathrm{C}^{-1}\right) \theta+(200 \mathrm{~g})\left(1 \mathrm{cal} \mathrm{g}^{-1}{ }^{\circ} \mathrm{C}^{-1}\right) \theta \text {. }
\)
Thus,
\(
\begin{aligned}
& 2 \cdot 4 \times 10^4 \mathrm{cal}=2400 \mathrm{cal}+16000 \mathrm{cal}+\left(220 \mathrm{cal}^{\circ} \mathrm{C}^{-1}\right) \theta \\
& \text { or, } \quad \theta=\frac{5600 \mathrm{cal}}{220 \mathrm{cal}^{\circ} \mathrm{C}^{-1}}=25 \cdot 5^{\circ} \mathrm{C} .
\end{aligned}
\)

Hint

Total mass of the water \(=M=100 \mathrm{~g}\).
Latent heat of vaporization of water at \(0^{\circ} \mathrm{C}\)
\(
=L_1=21.0 \times 10^5 \mathrm{~J} \mathrm{~kg}^{-1} \text {. }
\)
Latent heat of fusion of ice \(=L_2=3.36 \times 10^5 \mathrm{~J} \mathrm{~kg}^{-1}\).
Suppose, the mass of the ice formed \(=m\).
Then the mass of water evaporated \(=M-m\).
Heat taken by the water to evaporate \(=(M-m) L_1\) and heat given by the water in freezing \(=m L_2\).
Thus, \(m L_2=(M-m) L_1\)
or,
\(
\begin{aligned}
m & =\frac{M L_1}{L_1+L_2} \\
& =\frac{(100 \mathrm{~g})\left(21 \cdot 0 \times 10^5 \mathrm{~J} \mathrm{~kg}^{-1}\right)}{(21 \cdot 0+3.36) \times 10^5 \mathrm{~J} \mathrm{~kg}^{-1}}=86 \mathrm{~g} .
\end{aligned}
\)

Hint

Let the mass of the bullet \(=m\).
Heat required to take the bullet from \(27^{\circ} \mathrm{C}\) to \(327^{\circ} \mathrm{C}\)
\(
\begin{aligned}
& =m \times\left(125 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}\right)(300 \mathrm{~K}) \\
& =m \times\left(3.75 \times 10^4 \mathrm{~J} \mathrm{~kg}^{-1}\right) .
\end{aligned}
\)
Heat required to melt the bullet
\(
=m \times\left(2.5 \times 10^4 \mathrm{~J} \mathrm{~kg}^{-1}\right) \text {. }
\)
If the initial speed be \(v\), the kinetic energy is \(\frac{1}{2} m v^2\) and hence the heat developed is \(\frac{1}{2}\left(\frac{1}{2} m v^2\right)=\frac{1}{4} m v^2\). Thus,
\(
\frac{1}{4} m v^2=m(3 \cdot 75+2 \cdot 5) \times 10^4 \mathrm{~J} \mathrm{~kg}^{-1}
\)
or, \(\quad v=500 \mathrm{~m} \mathrm{~s}^{-1}\).

Hint

The initial gravitational potential energy of the ball
\(
\begin{aligned}
& =m g h \\
= & m \times(10 \mathrm{~m} \mathrm{~s} \\
= & m \times\left(6.2 \times 10^{-2} \mathrm{~m}^2 \mathrm{~s}^{-2}\right)=m \times\left(6.2 \times 10^4 \mathrm{~J} \mathrm{~kg}^{-1}\right) .
\end{aligned}
\)
All this energy is used to heat the ball as it reaches the ground with a small velocity. Energy required to take the ball from \(30^{\circ} \mathrm{C}\) to \(330^{\circ} \mathrm{C}\) is
\(
\begin{aligned}
& m \times\left(126 \mathrm{~J} \mathrm{~kg}^{-1}{ }^{\circ} \mathrm{C}^{-1}\right) \times\left(300^{\circ} \mathrm{C}\right) \\
= & m \times 37800 \mathrm{~J} \mathrm{~kg}^{-1}
\end{aligned}
\)
and energy required to melt the ball at \(330^{\circ} \mathrm{C}\)
\(
=m L
\)
where \(L=\) latent heat of fusion of lead.
Thus,
\(
m \times\left(6.2 \times 10^4 \mathrm{~J} \mathrm{~kg}^{-1}\right)=m \times 37800 \mathrm{~J} \mathrm{~kg}^{-1}+m L
\)
or, \(\quad L=2.4 \times 10^4 \mathrm{~J} \mathrm{~kg}^{-1}\).

Hint

The speed of sound in a gas is given by

\(
\begin{aligned}
v & =\sqrt{\frac{\gamma P}{\rho}}=\sqrt{\frac{\gamma R T}{M}} \\
\gamma & =\frac{M v^2}{R T} \\
& =\frac{\left(32 \times 10^{-3} \mathrm{~kg} \mathrm{~mol}^{-1}\right)\left(315 \mathrm{~m} \mathrm{~s}^{-1}\right)^2}{\left(8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)(273 \mathrm{~K})}=1.4
\end{aligned}
\)

Hint

For a diatomic gas, \(C_V=\frac{5}{2} R\) and \(C_p=\frac{7}{2} R\). The work done in an isobaric process is
\(
\begin{aligned}
W & =p\left(V_2-V_1\right) \\
& =n R T_2-n R T_1 \\
\text { or, } \quad T_2-T_1 & =\frac{W}{n R} .
\end{aligned}
\)
The heat given in an isobaric process is
\(
\begin{aligned}
Q & =n C_p\left(T_2-T_1\right) \\
& =n C_p \frac{W}{n R}=\frac{7}{2} W \\
& =\frac{7}{2} \times 200 \mathrm{~J}=700 \mathrm{~J}
\end{aligned}
\)

Hint

Let,
\(C_V^{\prime}=\) molar heat capacity of the first gas,
\(C^{\prime \prime}{ }_V=\) molar heat capacity of the second gas,
\(C_V=\) molar heat capacity of the mixture
and similar symbols for other quantities. Then,
\(
\gamma=\frac{C_p^{\prime}}{C_V^{\prime}}=1.67
\)
and \(\quad C_p^{\prime}=C_V^{\prime}+R\).
This gives \(C_V^{\prime}=\frac{3}{2} R\) and \(C_p^{\prime}=\frac{5}{2} R\).
Similarly, \(\gamma=1 \cdot 4\) gives \(C^{\prime \prime}{ }_V=\frac{5}{2} R\) and \(C^{\prime \prime}{ }_p=\frac{7}{2} R\).
Suppose the temperature of the mixture is increased by \(d T\). The increase in the internal energy of the first gas \(=n_1 C_V{ }^{\prime} d T\). The increase in internal energy of the second gas \(=n_2 C_V^{\prime \prime} d T\) and the increase in internal energy of the mixture \(=\left(n_1+n_2\right) C_V d T\). Thus,
\(
\left(n_1+n_2\right) C_V d T=n_1 C_V^{\prime} d T+n_2 C_V^{\prime \prime} d T
\)
or,
\(
\begin{aligned}
C_V & =\frac{n_1 C_V^{\prime}+n_2 C_V^{\prime \prime}}{n_1+n_2} \dots(i) \\
C_p=C_V+R & =\frac{n_1 C_V^{\prime}+n_2 C_V^{\prime \prime}}{n_1+n_2}+R \\
& =\frac{n_1\left(C_V^{\prime}+R\right)+n_2\left(C_V^{\prime \prime}+R\right)}{n_1+n_2} \\
& =\frac{n_1 C_p^{\prime}+n_2 C_p^{\prime \prime}}{n_1+n_2} \dots(ii)
\end{aligned}
\)
From (i) and (ii), \(\gamma=\frac{C_p}{C_V}=\frac{n_1 C_p^{\prime}+n_2 C_p^{\prime \prime}}{n_1 C_V^{\prime}+n_2 C_V^{\prime \prime}}\)
\(
=\frac{4 \times \frac{5}{2} R+2 \times \frac{7}{2} R}{4 \times \frac{3}{2} R+2 \times \frac{5}{2} R}=1.54
\)

Hint

(a) For an adiabatic process,
\(
p_1 V_1^\gamma=p_2 V_2^\gamma \text {. }
\)
Thus,
\(
\begin{gathered}
p_2=p_1\left(\frac{V_1}{V_2}\right)^\gamma \\
=(150 \mathrm{kPa})\left(\frac{1600 \mathrm{~cm}^3}{400 \mathrm{~cm}^3}\right)^{3 / 2}=1200 \mathrm{kPa} .
\end{gathered}
\)
(b) Work done by the gas in an adiabatic process is
\(
\begin{aligned}
W & =\frac{p_1 V_1-p_2 V_2}{\gamma-1} \\
& =\frac{(150 \mathrm{kPa})\left(1600 \mathrm{~cm}^3\right)-(1200 \mathrm{kPa})\left(400 \mathrm{~cm}^3\right)}{1 \cdot 5-1} \\
& =\frac{240 \mathrm{~J}-480 \mathrm{~J}}{0 \cdot 5}=-480 \mathrm{~J} .
\end{aligned}
\)

Hint

(a) When the gas is compressed in a short time, the process is adiabatic. Thus,
or,
\(
\begin{aligned}
T_2 V_2^{\gamma-1} & =T_1 V_1^{\gamma-1} \\
T_2 & =T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1} \\
& =(300 \mathrm{~K}) \times\left[\frac{800}{200}\right]^{0.4}=522 \mathrm{~K} .
\end{aligned}
\)
Rise in temperature \(=T_2-T_1=222 \mathrm{~K}\).
(b) When the gas is compressed in a long time, the process is isothermal. Thus, the temperature remains equal to the temperature of the surrounding that is \(27^{\circ} \mathrm{C}\). The rise in temperature \(=0\).

Hint

We have \(p V=n R T\) or, \(n=\frac{p V}{R T}\). The amount of the gas in the reservoir is \(n_1=\frac{p_1 V}{R T}\) before the gas is taken out and \(n_2=\frac{p_2 V}{R T}\) after the gas is taken out. The amount taken out is
\(
\begin{aligned}
& \Delta n=n_1-n_2=\left(p_1-p_2\right) \frac{V}{R T} \\
= & \frac{(600-525) \times 10^3 \mathrm{~N} \mathrm{~m}^{-2} \times\left(40 \times 10^{-2} \mathrm{~m}\right)^3}{\left(8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right) \times(300 \mathrm{~K})} \\
= & 1.925 \mathrm{~mol} .
\end{aligned}
\)
The gas is heated to \(100^{\circ} \mathrm{C}\) and cools down as it passes through the calorimeter. The average final temperature of the gas is \(\frac{20^{\circ} \mathrm{C}+30^{\circ} \mathrm{C}}{2}=25^{\circ} \mathrm{C}\). Thus, the average decrease in temperature of the gas is
\(
\begin{aligned}
& \Delta T=\left(100^{\circ} \mathrm{C}-25^{\circ} \mathrm{C}\right)=75^{\circ} \mathrm{C} \\
& \Delta T=75 \mathrm{~K} .
\end{aligned}
\)
\(
\text { or, } \quad \Delta T=75 \mathrm{~K} \text {. }
\)
The heat lost by the gas is
\(
\Delta Q=\Delta n C_p \Delta T \text {. }
\)
The heat gained by the calorimeter and its contents is \((100 \mathrm{~g})\left(4200 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}\right)(30-20)^{\circ} \mathrm{C}=4200 \mathrm{~J}\).
Thus, \(\Delta n C_p \Delta T=4200 \mathrm{~J}\)
or,
\(
C_p=\frac{4200 \mathrm{~J}}{(1 \cdot 925 \mathrm{~mol})(75 \mathrm{~K})}=29 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \text {. }
\)

Hint

Suppose the sample contains \(n\) moles. Also, suppose the volume changes from \(V_1\) to \(V_2\) and the temperature changes from \(T_1\) to \(T_2\).
The heat supplied is
\(
\Delta Q=n C_p\left(T_2-T_1\right) .
\)
(a) The change in internal energy is
\(
\begin{aligned}
\Delta U & =n C_V\left(T_2-T_1\right)=\frac{C_V}{C_p} n C_p\left(T_2-T_1\right) \\
& =\frac{C_V}{C_p} \Delta Q=\frac{140 \mathrm{~J}}{1 \cdot 4}=100 \mathrm{~J} .
\end{aligned}
\)
(b) The work done by the gas is
\(
\begin{aligned}
\Delta W & =\Delta Q-\Delta U \\
& =140 \mathrm{~J}-100 \mathrm{~J}=40 \mathrm{~J} .
\end{aligned}
\)

Hint

We have
\(
C_V=C_p-R=2.5 R-R=1.5 R .
\)
The amount of the gas (in moles) is \(n=\frac{p V}{R T}\)
\(
=\frac{\left(1.6 \times 10^6 \mathrm{~N} \mathrm{~m}^{-2}\right) \times\left(0.0083 \mathrm{~m}^3\right)}{\left(8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)(300 \mathrm{~K})}=5.3 \mathrm{~mol} \text {. }
\)
As the gas is kept in a closed vessel, its volume is constant. Thus, we have
\(
\begin{aligned}
\Delta Q & =n C_V \Delta T \\
\text { or, } \quad \Delta T & =\frac{\Delta Q}{n C_V} \\
& =\frac{2.49 \times 10^4 \mathrm{~J}}{(5.3 \mathrm{~mol})\left(1.5 \times 8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)}=377 \mathrm{~K} .
\end{aligned}
\)
The final temperature is \(300 \mathrm{~K}+377 \mathrm{~K}=677 \mathrm{~K}\).
We have,
\(
\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2} .
\)
Here \(V_1=V_2\) Thus,
\(
\begin{aligned}
p_2 & =\frac{T_2}{T_1} p_1=\frac{677}{300} \times 1.6 \times 10^6 \mathrm{~N} \mathrm{~m}^{-2} \\
& =3.6 \times 10^6 \mathrm{~N} \mathrm{~m}^{-2} .
\end{aligned}
\)

Hint

Using \(p V=n R T\), the volume of 1 mole of air at STP is
\(
\begin{aligned}
V & =\frac{n R T}{p}=\frac{(1 \mathrm{~mol}) \times\left(8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right) \times(273 \mathrm{~K})}{1.0 \times 10^6 \mathrm{~N} \mathrm{~m}^{-2}} \\
& =0.0224 \mathrm{~m}^3 .
\end{aligned}
\)
The mass of 1 mole is, therefore,
\(\left(1.29 \mathrm{~kg} \mathrm{~m}^{-2}\right) \times\left(0.0224 \mathrm{~m}^3\right)=0.029 \mathrm{~kg}\).
The number of moles in \(1 \mathrm{~kg}\) is \(\frac{1}{0 \cdot 029}\). The molar heat capacity at constant volume is
\(
\begin{aligned}
C_V & =\frac{170 \mathrm{cal}}{(1 / 0 \cdot 029) \mathrm{mol} \mathrm{K}^{-1}} \\
& =4.93 \mathrm{cal} \mathrm{K}^{-1} \mathrm{~mol}^{-1} .
\end{aligned}
\)
Hence, \(\quad C_p=\gamma C_V=1.4 \times 4.93 \mathrm{cal} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\)
or, \(\quad C_p-C_V=0.4 \times 4.93 \mathrm{cal} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\)
\(=1.97 \mathrm{cal} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\).
Also,
\(
C_p-C_V=R=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} .
\)
Thus, \(\quad 8.3 \mathrm{~J}=1.97 \mathrm{cal}\).
The mechanical equivalent of heat is
\(
\frac{8.3 \mathrm{~J}}{1.97 \mathrm{cal}}=4.2 \mathrm{~J} \mathrm{cal}^{-1} .
\)

Hint

The amount of heat transferred through the slab to the ice in one hour is
\(
\begin{aligned}
Q & =\left(4.8 \times 10^{-3} \mathrm{~kg}\right) \times\left(3.36 \times 10^5 \mathrm{~J} \mathrm{~kg}^{-1}\right) \\
& =4.8 \times 336 \mathrm{~J} .
\end{aligned}
\)
Using the equation
\(
\begin{aligned}
Q & =\frac{K A\left(\theta_1-\theta_2\right) t}{x}, \\
4.8 \times 336 \mathrm{~J} & =\frac{K\left(3600 \mathrm{~cm}^2\right)\left(100^{\circ} \mathrm{C}\right) \times(3600 \mathrm{~s})}{10 \mathrm{~cm}}
\end{aligned}
\)
or, \(\quad K=1.24 \times 10^{-3} \mathrm{~W} \mathrm{~m}^{-10} \mathrm{C}^{-1}\).

Hint

The situation is shown in the figure below. Let the temperature inside the box be \(\theta\). The rate at which heat enters the box through the left plug is
\(
\frac{\Delta Q_1}{\Delta t}=\frac{K A\left(\theta_1-\theta\right)}{x} .
\)
The rate of heat generation in the box \(=13 \mathrm{~W}\). The rate at which heat flows out of the box through the right plug is
\(
\frac{\Delta Q_2}{\Delta t}=\frac{K A\left(\theta-\theta_2\right)}{x} .
\)
In the steady state
\(
\begin{aligned}
\frac{\Delta Q_1}{\Delta t}+13 \mathrm{~W} & =\frac{\Delta Q_2}{\Delta t} \\
\text { or, } \quad \frac{K A}{x}\left(\theta_1-\theta\right)+13 \mathrm{~W} & =\frac{K A}{x}\left(\theta-\theta_2\right) \\
\text { or, } & 2 \frac{K A}{x} \theta=\frac{K A}{x}\left(\theta_1+\theta_2\right)+13 \mathrm{~W} \\
\text { or, } & \theta=\frac{\theta_1+\theta_2}{2}+\frac{(13 \mathrm{~W}) x}{2 K A} \\
= & \frac{100^{\circ} \mathrm{C}+4^{\circ} \mathrm{C}}{2}+\frac{(13 \mathrm{~W}) \times 0.08 \mathrm{~m}}{2 \times\left(2 \cdot 0 \mathrm{~W} \mathrm{~m}^{-1_{\circ}} \mathrm{C}^{-1}\right)\left(12 \times 10^{-4} \mathrm{~m}^2\right)} \\
= & 52^{\circ} \mathrm{C}+216.67^{\circ} \mathrm{C} \approx 269^{\circ} \mathrm{C} .
\end{aligned}
\)

Hint

The situation is shown in the figure below. Let the temperature at the junction be \(\theta\) (on Celsius scale). The same heat current passes through the copper and the steel rods. Thus,
\(
\begin{aligned}
& \frac{\Delta Q}{\Delta t}=\frac{K_{\text {cu }} A\left(100^{\circ} \mathrm{C}-\theta\right)}{75 \mathrm{~cm}}=\frac{K_{\text {steel }} A \theta}{125 \mathrm{~cm}} \\
& \frac{K_{c u}\left(100^{\circ} \mathrm{C}-\theta\right)}{75}=\frac{K_{\text {steel }} \theta}{125} \\
& \frac{100^{\circ} \mathrm{C}-\theta}{\theta}=\frac{75 K_{\text {steel }}}{125 K_{\text {cu }}}=\frac{3}{5} \times \frac{46}{386} \\
& \theta=93^{\circ} \mathrm{C} .
\end{aligned}
\)
The rate of heat flow is
\(
\begin{aligned}
\frac{\Delta Q}{\Delta t} & =\frac{K_{\text {steel }} A \theta}{125 \mathrm{~cm}} \\
& =\frac{\left(46 \mathrm{~J} \mathrm{~s}^{-1} \mathrm{~m}^{-1} \mathrm{C}^{-1}\right)\left(\pi \times 1 \mathrm{~cm}^2\right) \times 93^{\circ} \mathrm{C}}{125 \mathrm{~cm}} \\
& =1.07 \mathrm{~J} \mathrm{~s}^{-1} .
\end{aligned}
\)

Hint

(a) Let the temperature of the interface be \(\theta\). The area of cross section of each plate is \(A=100 \mathrm{~cm}^2\) \(=0.01 \mathrm{~m}^2\). The thicknesses are \(x_A=0.04 \mathrm{~m}\) and \(x_B=0.025 \mathrm{~m}\).
The thermal resistance of the plate \(A\) is
\(
R_1=\frac{1}{K_A} \frac{x_A}{A}
\)
and that of the plate \(B\) is
\(
R_2=\frac{1}{K_B} \frac{x_B}{A}
\)
The equivalent thermal resistance is
\(
R=R_1+R_2=\frac{1}{A}\left(\frac{x_A}{K_A}+\frac{x_B}{K_B}\right) \text {. } \dots(i)
\)
Thus, \(\frac{\Delta Q}{\Delta t}=\frac{\theta_1-\theta_2}{R}\)
\(
\begin{aligned}
& =\frac{A\left(\theta_1-\theta_2\right)}{x_A / K_A+x_B / K_B} \\
& =\frac{\left(0.01 \mathrm{~m}^2\right)\left(100^{\circ} \mathrm{C}\right)}{(0.04 \mathrm{~m}) /\left(200 \mathrm{~W} \mathrm{~m}^{-1_{\circ}} \mathrm{C}^{-1}\right)+(0.025 \mathrm{~m}) /\left(400 \mathrm{~W} \mathrm{~m}^{-1_{\circ}} \mathrm{C}^{-1}\right)} \\
& =3810 \mathrm{~W} .
\end{aligned}
\)
(b) We have \(\frac{\Delta Q}{\Delta t}=\frac{A\left(\theta-\theta_2\right)}{x_B / K_B}\)
or, \(\quad 3810 \mathrm{~W}=\frac{\left(0.01 \mathrm{~m}^2\right)\left(\theta-0^{\circ} \mathrm{C}\right)}{(0.025 \mathrm{~m}) /\left(400 \mathrm{~W} \mathrm{~m}^{-1 \circ} \mathrm{C}^{-1}\right)}\)
or, \(\theta=24^{\circ} \mathrm{C}\).
(c) If \(K\) is the equivalent thermal conductivity of the compound plate, its thermal resistance is
\(
R=\frac{1}{A} \frac{x_A+x_B}{K} .
\)
Comparing with (i),
\(
\begin{aligned}
\frac{x_A+x_B}{K} & =\frac{x_A}{K_A}+\frac{x_B}{K_B} \\
K & =\frac{x_A+x_B}{x_A / K_A+x_B / K_B} \\
& =248 \mathrm{~W} \mathrm{~m}^{-1_0} \mathrm{C}^{-1}
\end{aligned}
\)

Hint

The area of the roof
\(
=4 \mathrm{~m} \times 4 \mathrm{~m}=16 \mathrm{~m}^2 \text {. }
\)
The thickness \(x=10 \mathrm{~cm}=0 \cdot 10 \mathrm{~m}\).
(a) The thermal resistance of the roof is
\(
\begin{aligned}
R_1=\frac{1}{K} \frac{x}{A} & =\frac{1}{1 \cdot 26 \mathrm{~W} \mathrm{~m}^{-1} { }^{\circ} \mathrm{C}^{-1}} \frac{0 \cdot 10 \mathrm{~m}}{16 \mathrm{~m}^2} \\
& =4 \cdot 96 \times 10^{-8}{ }^{\circ} \mathrm{C} \mathrm{W}^{-1} .
\end{aligned}
\)
The heat current is
\(
\begin{aligned}
\frac{\Delta Q}{\Delta t} & =\frac{\theta_1-\theta_2}{R_1}=\frac{46^{\circ} \mathrm{C}-32^{\circ} \mathrm{C}}{4 \cdot 96 \times 10^{-3}{ }^{\circ} \mathrm{C} \mathrm{W}^{-1}} \\
& =2822 \mathrm{~W} .
\end{aligned}
\)
(b) The thermal resistance of the brick layer is
\(
\begin{aligned}
R_2=\frac{1}{K} \frac{x}{A}=\frac{1}{0.65 \mathrm{~W} \mathrm{~m}^{-1} { }^{\circ} \mathrm{C}^{-1}} \frac{7 \cdot 5 \times 10^{-2} \mathrm{~m}}{16 \mathrm{~m}^2}
& =7.2 \times 10^{-3}{ }^{\circ} \mathrm{C} \mathrm{W}^{-1} .
\end{aligned}
\)
The equivalent thermal resistance is
\(
\begin{aligned}
R & =R_1+R_2=(4 \cdot 96+7 \cdot 2) \times 10^{-3}{ }^{\circ} \mathrm{C} \mathrm{W}^{-1} \\
& =1.216 \times 10^{-2}{ }^{\circ} \mathrm{C} \mathrm{W}^{-1} .
\end{aligned}
\)
The heat current is
\(
\begin{aligned}
\frac{\Delta Q}{\Delta t} & =\frac{\theta_1-\theta_2}{R}=\frac{46^{\circ} \mathrm{C}-32^{\circ} \mathrm{C}}{1.216 \times 10^{-2}{ }^{\circ} \mathrm{C} \mathrm{W}^{-1}} . \\
& =1152 \mathrm{~W} .
\end{aligned}
\)

Hint

The situation is shown in figure above.
The thermal resistances of the wood, the cement and the brick layers are
\(
\begin{aligned}
R_W & =\frac{1}{K} \frac{x}{A} \\
& =\frac{1}{0.125 \mathrm{~W} \mathrm{~m}^{-1}{ }^{\circ} \mathrm{C}^{-1}} \frac{2.5 \times 10^{-2} \mathrm{~m}}{137 \mathrm{~m}^2} \\
& =\frac{0.20}{137}{ }^{\circ} \mathrm{C} \mathrm{W}^{-1}, \\
R_C & =\frac{1}{1.5 \mathrm{~W}{ }^{-1}{ }^{\circ} \mathrm{C}^{-1}} \frac{1.0 \times 10^{-2} \mathrm{~m}}{137 \mathrm{~m}^2} \\
& =\frac{0.0067}{137}{ }^{\circ} \mathrm{C} \mathrm{W}^{-1} \\
R_B & =\frac{1}{1.0 \mathrm{~W} \mathrm{~m}^{-1}{ }^{\circ} \mathrm{C}^{-1}} \frac{25.0 \times 10^{-2} \mathrm{~m}}{137 \mathrm{~m}^2} \\
& =\frac{0.25}{137}{ }^{\circ} \mathrm{C} \mathrm{W}^{-1} .
\end{aligned}
\)
As the layers are connected in series, the equivalent thermal resistance is
\(
R=R_W+R_C+R_B
\)
\(
\begin{aligned}
& =\frac{0.20+0.0067+0.25}{137}{ }^{\circ} \mathrm{C} \mathrm{W}^{-1} \\
& =3.33 \times 10^{-3}{ }^{\circ} \mathrm{C} \mathrm{W}^{-1} .
\end{aligned}
\)
The heat current is
\(
\begin{aligned}
i & =\frac{\theta_1-\theta_2}{R} \\
& =\frac{20^{\circ} \mathrm{C}-\left(-10^{\circ} \mathrm{C}\right)}{3.33 \times 10^{-3}{ }^{\circ} \mathrm{C} \mathrm{W}{ }^{-1}} \approx 9000 \mathrm{~W} .
\end{aligned}
\)
The heater must supply \(9000 \mathrm{~W}\) to compensate the outflow of heat.

Hint

It is clear from the symmetry of the figure that the points \(C\) and \(D\) are equivalent in all respect and hence, they are at the same temperature, say \(\theta\). No heat will flow through the rod \(C D\). We can, therefore, neglect this rod in further analysis.
Let \(l\) and \(A\) be the length and the area of cross section of each rod. The thermal resistances of \(A B, B C\) and \(B D\) are equal. Each has a value
\(R_1=\frac{1}{K_x} \frac{l}{A} .\)
Similarly, thermal resistances of \(C E\) and \(D E\) are equal, each having a value
\(R_2=\frac{1}{K_y} \frac{l}{A} .\)
As the rod \(C D\) has no effect, we can say that the rods \(B C\) and \(C E\) are joined in series. Their equivalent thermal resistance is
\(R_3=R_{B C}+R_{C Z}=R_1+R_2 \text {. }\)
Also, the rods \(B D\) and \(D E\) together have an equivalent thermal resistance \(R_4=R_{D D}+R_{D E}=R_1+R_2\).
The resistances \(R_3\) and \(R_4\) are joined in parallel and hence their equivalent thermal resistance is given by
\(\frac{1}{R_5}=\frac{1}{R_3}+\frac{1}{R_4}=\frac{2}{R_3}\)
or,
\(R_5=\frac{R_3}{2}=\frac{R_1+R_2}{2}\)
This resistance \(R_5\) is connected in series with \(A B\). Thus, the total arrangement is equivalent to a thermal resistance
\(R=R_{A B}+R_{\mathrm{b}}=R_1+\frac{R_1+R_2}{2}=\frac{3 R_1+R_2}{2}\)
The figure below shows the successive steps in this reduction.
The heat current through \(A\) is
\(i=\frac{\theta_A-\theta_E}{R}=\frac{2\left(\theta_A-\theta_E\right)}{3 R_1+R_2}\)
This current passes through the rod \(A B\). We have
or, \(\quad \theta_A-\theta_B=\left(R_{A D}\right) i\)
\(i=\frac{\theta_A-\theta_B}{R_{A B}}\)
\(=R_1 \frac{2\left(\theta_A-\theta_{\bar{E}}\right)}{3 R_1+R_2} \text {. }\)
Putting from (i) and (ii),
\(
\begin{aligned}
\theta_A-\theta_B & =\frac{2 K_y\left(\theta_A-\theta_{\bar{Z}}\right)}{K_x+3 K_y} \\
& =\frac{2 \times 400}{800+3 \times 400} \times 50^{\circ} \mathrm{C}=20^{\circ} \mathrm{C}
\end{aligned}
\)
or,
\(\theta_B=\theta_A-20^{\circ} \mathrm{C}=40^{\circ} \mathrm{C} \text {. }\)

Hint

The thermal resistance of \(A C\) is equal to that of \(C B\) and is equal to \(2.5 \mathrm{~K} \mathrm{~W}^{-1}\). Suppose, the temperature at \(C\) is \(\theta\). The heat currents through \(A C\), \(C B\) and \(C D\) are
\(
\frac{\Delta Q_1}{\Delta t}=\frac{100^{\circ} \mathrm{C}-\theta}{2.5 \mathrm{~K} \mathrm{~W}^{-1}},
\)
\(
\begin{array}{rlrl}
\frac{\Delta Q_2}{\Delta t} & =\frac{\theta-0^{\circ} \mathrm{C}}{2 \cdot 5 \mathrm{~K} \mathrm{~W}^{-1}} \\
\text { and } & \frac{\Delta Q_3}{\Delta t} =\frac{\theta-25^{\circ} \mathrm{C}}{5 \cdot 0 \mathrm{~K} \mathrm{~W}^{-1}}
\end{array}
\)
We also have
\(
\frac{\Delta Q_1}{\Delta t}=\frac{\Delta Q_2}{\Delta t}+\frac{\Delta Q_3}{\Delta t}
\)
Thus,
\(\begin{aligned} \frac{100^{\circ} \mathrm{C}-\theta}{2 \cdot 5} & =\frac{\theta-0^{\circ} \mathrm{C}}{2 \cdot 5}+\frac{\theta-25^{\circ} \mathrm{C}}{5} \\ 225^{\circ} \mathrm{C} & =5 \theta \\ \theta & =45^{\circ} \mathrm{C} .\end{aligned}\)
\(
\frac{\Delta Q_3}{\Delta t}=\frac{45^{\circ} \mathrm{C}-25^{\circ} \mathrm{C}}{5 \cdot 0 \mathrm{~K} \mathrm{~W}^{-1}}=\frac{20 \mathrm{~K}}{5 \cdot 0 \mathrm{~K} \mathrm{~W}^{-1}}
\)
\(
=4.0 \mathrm{~W} \text {. }
\)

Hint

Let us draw two spherical shells of radii \(x\) and \(x+d x\) concentric with the given system. Let the temperatures at these shells be \(\theta\) and \(\theta+d \theta\) respectively. The amount of heat flowing radially inward through the material between \(x\) and \(x+d x\) is
\(
\frac{\Delta Q}{\Delta t}=\frac{K 4 \pi x^2 d \theta}{d x} .
\)
Thus,
\(
\begin{aligned}
K 4 \pi \int_{\theta_1}^{\theta_2} d \theta & =\frac{\Delta Q}{\Delta t} \int_{r_1}^{r_2} \frac{d x}{x^2} \\
K 4 \pi\left(\theta_2-\theta_1\right) & =\frac{\Delta Q}{\Delta t}\left(\frac{1}{r_1}-\frac{1}{r_2}\right) \\
\frac{\Delta Q}{\Delta t} & =\frac{4 \pi K r_1 r_2\left(\theta_2-\theta_1\right)}{r_2-r_1} .
\end{aligned}
\)

Hint

Suppose, the ice starts forming at time \(t=0\) and a thickness \(x\) is formed at time \(t\). The amount of heat flown from the water to the surrounding in the time interval \(t\) to \(t+d t\) is
\(
\Delta Q=\frac{K A \theta}{x} d t .
\)
The mass of the ice formed due to the loss of this amount of heat is
\(
d m=\frac{\Delta Q}{L}=\frac{K A \theta}{x L} d t .
\)
The thickness \(d x\) of ice formed in time \(d t\) is
\(
\begin{aligned}
d x & =\frac{d m}{A \rho}=\frac{K \theta}{\rho x L} d t \\
\text { or, } \quad d t & =\frac{\rho L}{K \theta} x d x .
\end{aligned}
\)
Thus, the time \(T\) taken for the whole mass of water to freeze is given by
\(
\begin{gathered}
\int_0^T d t=\frac{\rho L}{K \theta} \int_0^h x d x \\
T=\frac{\rho L h^2}{2 K \theta} .
\end{gathered}
\)

Hint

Suppose, the temperature of the water in the smaller vessel is \(\theta\) at time \(t\). In the next time interval \(d t\), a heat \(\Delta Q\) is transferred to it where
\(
\Delta Q=\frac{K A}{L}\left(\theta_0-\theta\right) d t \dots(i)
\)
This heat increases the temperature of the water of mass \(m\) to \(\theta+d \theta\) where
\(
\Delta Q=m s d \theta \dots(ii)
\)
From (i) and (ii),
or,
\(
\frac{K A}{L}\left(\theta_0-\theta\right) d t=m s d \theta
\)
or,
\(
\begin{aligned}
d t & =\frac{L m s}{K A} \frac{d \theta}{\theta_0-\theta} \\
\int_0^T d t & =\frac{L m s}{K A} \int_{\theta_1}^{\theta_2} \frac{d \theta}{\theta_0-\theta}
\end{aligned}
\)
where \(T\) is the time required for the temperature of the water to become \(\theta_2\).
Thus,
\(
T=\frac{L m s}{K A} \ln \frac{\theta_0-\theta_1}{\theta_0-\theta_2}
\)

Hint

The initial temperature difference is \(97^{\circ} \mathrm{C}-5^{\circ} \mathrm{C}=92^{\circ} \mathrm{C}\) and at \(5.0 \mathrm{~s}\) the temperature difference becomes \(97^{\circ} \mathrm{C}-9^{\circ} \mathrm{C}=88^{\circ} \mathrm{C}\). As the change in the temperature difference is small, we work with the average temperature difference
\(
\frac{92^{\circ} \mathrm{C}+88^{\circ} \mathrm{C}}{2}=90^{\circ} \mathrm{C}=90 \mathrm{~K} .
\)
The rise in the temperature of the gas is
\(
9.0^{\circ} \mathrm{C}-5.0^{\circ} \mathrm{C}=4^{\circ} \mathrm{C}=4 \mathrm{~K} \text {. }
\)
The heat supplied to the gas in \(5.0 \mathrm{~s}\) is
\(
\begin{aligned}
\Delta Q & =n C_v \Delta T \\
& =(1 \mathrm{~mol}) \times\left(\frac{3}{2} \times 8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right) \times(4 \mathrm{~K}) \\
& =49 \cdot 8 \mathrm{~J} .
\end{aligned}
\)
If the thermal conductivity is \(K\),
\(
\begin{aligned}
49.8 \mathrm{~J} & =\frac{K\left(800 \times 10^{-4} \mathrm{~m}^2\right) \times(90 \mathrm{~K})}{1.0 \times 10^{-2} \mathrm{~m}} \times 5.0 \mathrm{~s} \\
\text { or, } \quad K & =\frac{49.8 \mathrm{~J}}{3600 \mathrm{msK}}=0.014 \mathrm{~J} \mathrm{~s}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1} .
\end{aligned}
\)

Hint

As the volume of the gas is constant, a heat \(\Delta Q\) given to the gas increases its temperature by \(\Delta T=\Delta Q / C_v\). Also, for a monatomic gas, \(C_v=\frac{3}{2} R\). If the temperature of the gas at time \(t\) is \(T\), the heat current into the gas is
\(
\begin{array}{rlrl}
\frac{\Delta Q}{\Delta t}  =\frac{K A\left(T_s-T\right)}{x} \\
\text { or, } & \frac{\Delta T}{\Delta t} =\frac{2 K A}{3 x R}\left(T_s-T\right) \\
\text { or, } & \int_{T_0}^T \frac{d T}{T_s-T} =\int_0^t \frac{2 K A}{3 x R} d t \\
\text { or, } & \ln \frac{T_s-T_0}{T_s-T} =\frac{2 K A}{3 x R} t \\
\text { or, } & T_s-T =\left(T_s-T_0\right) e^{-\frac{2 K A}{3 x R} t} \\
\text { or, } & T=T_s =\left(T_s-T_0\right) e^{-\frac{2 K A}{3 x R} t} .
\end{array}
\)
As the volume remains constant,
\(
\begin{aligned}
\frac{p}{T} & =\frac{p_0}{T_0} \\
\text { or, } \quad p & =\frac{p_0}{T_0} T \\
& =\frac{p_0}{T_0}\left[T_s-\left(T_s-T_0\right) e^{-\frac{2 K A}{3 x_R} t}\right] .
\end{aligned}
\)

Hint

The area of the blackbody is \(A=10^{-4} \mathrm{~m}^2\), its temperature is \(T_1=327^{\circ} \mathrm{C}=600 \mathrm{~K}\) and the temperature of the enclosure is \(T_2=27^{\circ} \mathrm{C}=300 \mathrm{~K}\). The blackbody emits radiation at the rate of \(A \sigma T_1^4\). The radiation falls on it (and gets absorbed) at the rate of \(A \sigma T_2^4\). The net rate of loss of energy is \(A \sigma\left(T_1^4-T_2^4\right)\). The heater must supply this much of power. Thus, the power needed is \(A \sigma\left(T_1^4-T_2^4\right)\) \(=\left(10^{-4} \mathrm{~m}^2\right)\left(6.0 \times 10^{-8} \mathrm{~W} \mathrm{~m}^{-2} \mathrm{~K}^{-4}\right)\left[(600 \mathrm{~K})^4-(300 \mathrm{~K}){ }^4\right]\) \(=0.73 \mathrm{~W}\).

Hint

Let the temperature of the coil be \(T\). The coil will emit radiation at a rate \(A \sigma T^4\). Thus,
\(
\begin{aligned}
1000 \mathrm{~W} & =\left(0.020 \mathrm{~m}^2\right) \times\left(6.0 \times 10^{-8} \mathrm{~W} \mathrm{~m}^{-2} \mathrm{~K}^{-4}\right) \times T^4 \\
\text { or, } \quad T^4 & =\frac{1000}{0.020 \times 6.00 \times 10^{-8}} \mathrm{~K}^4 \\
& =8.33 \times 10^{11} \mathrm{~K}^4 \\
\text { or, } \quad T & =955 \mathrm{~K} .
\end{aligned}
\)

Hint

Let the diameter of the sun be \(D\) and its distance from the earth be \(R\). From the question,
\(
\begin{aligned}
\frac{D}{R} & \approx 0.53 \times \frac{\pi}{180} \\
& =9.25 \times 10^{-3} \dots(i)
\end{aligned}
\)
The radiation emitted by the surface of the sun per unit time is
\(
4 \pi\left(\frac{D}{2}\right)^2 \sigma T^4=\pi D^2 \sigma T^4 .
\)
At distance \(R\), this radiation falls on an area \(4 \pi R^2\) in unit time. The radiation received at the earth’s surface per unit time per unit area is, therefore,
\(
\begin{aligned}
& \frac{\pi D^2 \sigma T^4}{4 \pi R^2}=\frac{\sigma T^4}{4}\left(\frac{D}{R}\right)^2 \\
& \text { Thus, } \quad \frac{\sigma T^4}{4}\left(\frac{D}{R}\right)^2=8.2 \mathrm{~J} \mathrm{~cm}^{-2} \mathrm{~min}^{-1} \\
& \text { or, } \quad \frac{1}{4} \times\left(5.67 \times 10^{-8} \mathrm{~W} \mathrm{~m}^{-2} \mathrm{~K}^{-4}\right) T^4 \times\left(9.25 \times 10^{-3}\right)^2 \\
&=\frac{8.2}{10^{-4} \times 60} \mathrm{~W} \mathrm{~m}^{-2} \\
& \text { or, } \quad T=5794 \mathrm{~K} \approx 5800 \mathrm{~K} .
\end{aligned}
\)

Hint

As the temperature differences are small, we can use Newton’s law of cooling.
or, \(\quad \frac{d \theta}{\theta-\theta_0}=-k d t\)
where \(k\) is a constant, \(\theta\) is the temperature of the body at time \(t\) and \(\theta_0=16^{\circ} \mathrm{C}\) is the temperature of the surrounding. We have,
\(
\begin{aligned}
& \int_{40^{\circ} \mathrm{C}}^{36 \mathrm{C}} \frac{d \theta}{\theta-\theta_0}=-k(5 \mathrm{~min}) \\
& \text { or, } \quad \ln \frac{36^{\circ} \mathrm{C}-16^{\circ} \mathrm{C}}{40^{\circ} \mathrm{C}-16^{\circ} \mathrm{C}}=-k(5 \min ) \\
& \text { or, } \quad k=-\frac{\ln (5 / 6)}{5 \mathrm{~min}} \text {. } \\
&
\end{aligned}
\)
If \(t\) be the time required for the temperature to fall from \(36^{\circ} \mathrm{C}\) to \(32^{\circ} \mathrm{C}\) then by (i),
\(
\begin{aligned}
\int_{36^{\circ} \mathrm{C}}^{32^{\circ} \mathrm{C}} \frac{d \theta}{\theta-\theta_0} & =-k t \\
\text { or, } \quad \ln \frac{32^{\circ} \mathrm{C}-16^{\circ} \mathrm{C}}{36^{\circ} \mathrm{C}-16^{\circ} \mathrm{C}} & =\frac{\ln (5 / 6) t}{5 \min } \\
\text { or, } \quad t & =\frac{\ln (4 / 5)}{\ln (5 / 6)} \times 5 \mathrm{~min} \\
& =6.1 \mathrm{~min} .
\end{aligned}
\)

Hint

We have,
\(
\frac{d \theta}{d t}=-k\left(\theta-\theta_0\right)
\)
where \(\theta_0\) is the temperature of the surrounding and \(\theta\) is the temperature of the body at time \(t\). Suppose \(\theta=\theta_1\) at \(t=0\).
Then,
or,
\(
\begin{aligned}
& \int_{\theta_1}^\theta \frac{d \theta}{\theta-\theta_0} & =-k \int_0^t d t \\
\text { or, } & \ln \frac{\theta-\theta_0}{\theta_1-\theta_0} & =-k t \\
\text { or, } & \theta-\theta_0 & =\left(\theta_1-\theta_0\right) e^{-k t} .
\end{aligned}
\)
The body continues to lose heat till its temperature becomes equal to that of the surrounding. The loss of heat in this entire period is
\(
\Delta Q_m=m s\left(\theta_1-\theta_0\right) .
\)
This is the maximum heat the body can lose. If the body loses half this heat, the decrease in its temperature will be,
\(
\frac{\Delta Q_m}{2 m s}=\frac{\theta_1-\theta_0}{2} .
\)
If the body loses this heat in time \(t_1\), the temperature at \(t_1\) will be
\(
\theta_1-\frac{\theta_1-\theta_0}{2}=\frac{\theta_1+\theta_0}{2} .
\)
Putting these values of time and temperature in (i),
\(
\begin{aligned}
& \frac{\theta_1+\theta_0}{2}-\theta_0=\left(\theta_1-\theta_0\right) e^{-k t_1} \\
& \text { or, } \quad e^{-k t_1}=\frac{1}{2} \\
&
\end{aligned}
\)
or,
\(t_1=\frac{\ln 2}{k}\).

Hint

As the gas can leak out of the hole, the pressure inside the vessel will be equal to the atmospheric pressure \(p_a\). Let \(n\) be the amount of the gas (moles) in the vessel at time \(t\). Suppose an amount \(\Delta Q\) of heat is given to the gas in time \(d t\). Its temperature increases by \(d T\) where
\(
\Delta Q=n C_p d T
\)
If the temperature of the gas is \(T\) at time \(t\), we have
\(
\frac{\Delta Q}{d t}=\frac{T_a-T}{r}
\)
or, \(\quad\left(C_p r\right) n d T=\left(T_a-T\right) d t \dots(i)\).
We have, \(\quad p_a a^3=n R T\)
or, \(\quad n d T+T d n=0\)
or, \(\quad n d T=-T d n \dots(ii)\).
Also, \(\quad T=\frac{p_a a^3}{n R} \dots(iii)\).
Using (ii) and (iii) in (i),
\(
\begin{aligned}
& \frac{-C_p r p_a a^3}{n R} d n=\left(T_a-\frac{p_a a^3}{n R}\right) d t \\
& \text { or, } \quad \frac{d n}{n R\left(T_a-\frac{p_a a^3}{n R}\right)}=-\frac{d t}{C_p r p_a a^3} \\
& \text { or, } \quad \int_{n_0}^n \frac{d n}{n R T_a-p_a a^3}=-\int_0^t \frac{d t}{C_p r p_a a^3}
\end{aligned}
\)
where \(n_0=\frac{p_a a^3}{R T_0}\) is the initial amount of the gas in the vessel. Thus,
\(
\begin{aligned}
& \qquad \frac{1}{R T_a} \ln \frac{n R T_a-p_a a^3}{n_0 R T_a-p_a a^3}=-\frac{t}{C_p r p_a a^3} \\
& \text { or, } \quad n R T_a-p_a a^3=\left(n_0 R T_a-p_a a^3\right) e^{-\frac{R T_a}{C_p r p_a a^3} t} . \\
& \text { Writing } \quad n_0=\frac{p_a a^3}{R T_0} \text { and } C_p=C_v+R=\frac{7 R}{2}, \\
& n=\frac{p_a a^3}{R T_a}\left[1+\left(\frac{T_a}{T_0}-1\right) e^{-\frac{2 T_a}{7 r p_a a^3} t}\right] .
\end{aligned}
\)