HT MCQs

Hint

\(
\frac{\Delta Q}{\Delta t}=-k A \frac{d T}{d x}
\)
Here, the thermal conductivity \((k)\) depends on the material, and not the dimensions and mass of the body.

Hint

In conduction, heat is transferred from one place to other by the vibration of the molecules. In this process, the average position of a molecule does not change. Hence, there is no mass movement of matter.

In convection, heat is transferred from one place to other by the actual motion of particles of the medium. When water is heated, hot water moves upwards and cool water moves downwards.
In radiation process, the transfer of heat does not require any material medium.
For a room containing air, heat can be transferred via radiation (no medium required) and convection (by the movement of air molecules) and by conduction (due to the collision of hot air molecules with other molecules).

Hint

From Stefan-Boltzmann law, the energy of thermal radiation emitted per unit time by a blackbody of surface area \(A\) is given by \(u=\sigma A T^4\)
Here, \(\sigma \sigma\) is the Stefan-Boltzmann constant.
Since the temperature of the solid is less than the surroundings, the temperature of the solid will increase with time until it reaches equilibrium with the surroundings. The rate of emission from the solid will be proportional to \(T_1^4\) and rate of emission from the surroundings will be proportional to \(T_2^4\).
So, the net rate of increase in temperature will be proportional to \(T_2^4-T_1^4\).

Hint

From Stefan-Boltzmann law, the energy of the thermal radiation emitted per unit time by a blackbody of surface area \(A\) is given by \(u=\sigma A T^4\) here, \(\sigma\) is Stefan-Boltzmann constant. This law holds true for all the bodies.

Hint

From Stefan-Boltzmann law, energy of the thermal radiation emitted per unit time by a blackbody of surface area \(A\) is given by \(u=\sigma A T^4\)
Here, \(\sigma\) is Stefan-Boltzmann constant.
\(
\begin{aligned}
\frac{u A}{u B} & =\frac{\mathrm{T}_{\mathrm{A}}^4}{T_B^4} \\
\frac{u_A}{u_B} & =\frac{(273+10)^4}{(273+20)^4}= 1/1.15
\end{aligned}
\)

Hint

In a steady state, the temperature of the rod is nonuniform maximum at the end near the furnace and minimum at the end that is away from the furnace.

Hint

From Stefan-Boltzman’s law, the energy of the thermal radiation emitted per unit time by a blackbody of surface area \(A\) is given by, \(u=\sigma A T^4\)
Where \(\sigma\) is Stefan’s constant.
Suppose a body at temperature \(T\) is kept in a room at temperature \(T_0\).
According to Stefan’s law, energy of the thermal radiation emitted by the body per unit time is \(u=e \sigma A T^4\) Here, \(e\) is the emissivity of the body.
The energy absorbed per unit time by the body is (due to the radiation emitted by the walls of the room \(\Delta u=e \sigma A T_0^4\)
Thus, the net loss of thermal energy per unit time is
\(\Delta u=e \sigma A\left(T^4-T_0^4\right) \dots(i)\)
Newton law of cooling is given by
\(
\frac{d T}{d t}=-b A\left(T-T_0\right)
\)
This can be obtained from equation (i) by considering the temperature difference to be small and doing the binomial expansion.

Hint

When a hot liquid is kept in a big room, the liquid will loose its temperature with time. The thermal energy emitted by the liquid will be gained by the walls of the room. As the room is big, we can assume that the temperature difference between the room and the liquid is large. From Stephen’s law, the liquid emits thermal energy in proportion to \(T^4\), where \(T\) is the initial temperature of the liquid.
As the temperature decreases, the rate of loss of thermal energy will also decrease. So, the slope of the curve will also decrease. Therefore, the plot of temperature with time is best represented by the curve (a).

Hint

When a hot liquid is kept in a big room, then the liquid will lose temperature with time. The thermal energy emitted by the liquid will be gained by the walls of the room. As the room is big, we can assume that the temperature difference between the room and the liquid is large. From Stephen’s law, the liquid emits thermal energy in proportion to \(T^4\), where \(T\) is the initial temperature of the liquid. As the temperature decreases, the rate of loss will also decrease. So, the slope of the curve will also decrease. Finally, at equilibrium, the temperature of the room will become equal to the new temperature of the liquid. So, in steady state, the difference between the temperatures of the two will become zero.
A graph is plotted between the logarithm of the numerical value of the temperature difference between the liquid and the room is plotted against time. The logarithm converts the fourth power dependence into a linear dependence with some coefficient (property of log). So, the plot satisfying all the above properties will be a straight line.

Hint

Let the temperature of the surrounding be \(T^{\circ} \mathrm{CAverage}\) temperature of the liquid in first case \(=62.5^{\circ}\mathrm{C}\)
Average temperature difference from the surroundings \(=(62.5-T)^{\circ} \mathrm{C}\)
From newton’s law of cooling,
\(
\begin{aligned}
& 1^{\circ} \mathrm{C} \min ^{-1}=-\mathrm{bA}(62.5-\mathrm{T})^{\circ} \mathrm{C} \\
& \Rightarrow-b A=\frac{1}{62.5-t} {\min} ^{-1} \dots(i)
\end{aligned}
\)
For the second case,
Average temperature \(=57.5^{\circ} \mathrm{C}\)
Temperature difference from the surroundings \(=(57.5-7)^{\circ} \mathrm{C}\)
From Newton’s law of cooling and equation (i),
\(
\begin{aligned}
& \frac{5^{\circ} \mathrm{C}}{t}=-b A(57.5-T)^{\circ} \mathrm{C} \\
& \Rightarrow \frac{5^{\circ} \mathrm{C}}{t}=\frac{1}{62.5-t}(57.5-T)^{\circ} \mathrm{C} \\
& \Rightarrow t=\frac{5(62.5-T)}{(57.5-T)} \\
& \Rightarrow t>5 \text { minutes }
\end{aligned}
\)

Hint

The heat transfer will take place from the hot end to the cold end of the rod via conduction. So, with time, the temperature of the rod will increase from the end dipped in boiling water to the end dipped in melting, until it comes in equilibrium with its surroundings. In steady state, the temperature of the rod is nonuniform and constant, maximum at the end dipped in boiling water and minimum at the end dipped in melting ice. Equilibrium means that the system is stable. So, all the macroscopic variables describing the system will not change with time. Hence, the temperature of the rod will become constant once equilibrium is reached, but its value is different at different positions of the rod.

Hint

(c) reflect radiation
(d) refract radiation
A black body is an ideal concept. A black body is the one that absorbs all the radiation incident on it. So, a black body does not reflect and refract radiation.

Hint

Convection current is the movement of air (or any fluid) due to the difference in the temperatures. During summer days, there is temperature difference of air above the land and river. Due to this, a convection current is set from the river to the land during daytime. On the other hand, during night, a convection current is set from the land to the river. Therefore, a mild air always flows on the shore of a calm river due to the convection current.

Hint

(c) If both are picked up by bare hands, the steel will be felt hotter than the charcoal
(d) If the two are picked up from the lawn and kept in a cold chamber, the charcoal will lose heat at a faster rate than the steel.

In steel, conductivity is higher than charcoal. So, if both are picked up by bare hands, then heat transfer from the body (steel or charcoal) to our hand will be larger in the case of steel. Hence, steel will be hotter than charcoal.

On the other hand, the emissivity of charcoal is higher as compared to steel. So, if the two are picked up from the lawn and kept in a cold chamber, charcoal will lose heat at a faster rate than steel.

Hint

(a) the maximum intensity of radiation will be near the frequency \(2 v_0\)
(c) the total energy emitted will increase by a factor of 16
From Wein’s displacement law,
\(
\lambda \mathrm{mT}=\mathrm{b} \text { ( a constant) }
\)
or \(\frac{c T}{V m}=b\)
Here, \(T\) is the absolute temperature of the body.
So, as the temperature is doubled to keep the product on the left-hand side constant, frequency is also doubled.
From Stefan’s law, we know that the rate of energy emission is proportional to \(\mathrm{T}^4\)
This implies that total energy emitted will increase by a factor of \((2)^4\), which is equal to 16.

Hint

(a) Both will emit equal amount of radiation per unit time in the beginning.
(b) Both will absorb equal amount of radiation from the surrounding in the beginning.
Let the temperature of the surroundings be \(T_0\)
From the Stefan-Boltzmann law, the energy of thermal radiation emitted per unit time by a blackbody of surface area \(A\) is given by \(u=\sigma A T^4\)
Here, \(\sigma\) is Stephen’s constant.
Also, the energy absorbed per unit time by the body is given by
\(
u_0=e \sigma A T_0^4
\)
As the two spheres have equal radii and temperatures, their rate of absorption and emission will be equal in the beginning.

Hint

Given:
Thermal conductivity of the material, \(k=0.80 \mathrm{~W} \mathrm{~m}^{-1}{ }^{\circ} \mathrm{c}^{-1}\)
Area of the cross section of the slab, \(A=100 \mathrm{~cm}^2=10^{-2} \mathrm{~m}^2\)
Thickness of the slab, \(\Delta \mathrm{x}=1 \mathrm{~cm}=10^{-2} \mathrm{~m}\)
\(
\begin{aligned}
& \text { Rate of flow of heat }=\frac{\text { Temperature difference }}{\text { Thermal resisstance }} \\
& \Rightarrow \frac{\Delta Q}{\Delta t}=\frac{\Delta T}{\frac{\Delta x}{k A}} \\
& \frac{\Delta Q}{\Delta t}=\frac{(90-10) k . A}{\Delta x} \\
& \frac{\Delta Q}{\Delta t}=\frac{(80) \times 0.8 \times 10^{-2}}{\Delta x} 1 \\
& \frac{\Delta Q}{\Delta t}=64 J / s \\
& \frac{\Delta Q}{\Delta t}=64 \times 60=3840 \mathrm{~J} / \mathrm{min}
\end{aligned}
\)

Hint

\(
\text { Rate of flow of heat }=\frac{\text { Temperature differences }}{\text { Thermal resistance }}
\)
Thickness of the container, \(l=1 \mathrm{~cm}=10^{-2} \mathrm{~m}\)
Thermal conductivity of the styrofoam sheet, \(\mathrm{k}=0.025 \mathrm{~J} \mathrm{~s}^{-1} \mathrm{~m}^{-1}{ }^{\circ} \mathrm{C}^{-1}\)
Area, \(A=0.80 \mathrm{~m}^2\)
Thermal resistance, \(\frac{l}{K A}=\frac{10^2}{0.025 \times 0.80}\)
Temperature difference, \(\Delta T=T_1-T_2=300-80=220 \mathrm{~K}\)
Rate of flow of heat, \(\left(\frac{\Delta Q}{\Delta t}\right)=\frac{T 1-T 2}{\frac{l}{K} A}\)
\(
\Rightarrow\left(\frac{\Delta Q}{\Delta t}\right)=440 \mathrm{~J} / \mathrm{s}=440 W
\)

Hint

Rate of how of heat \(=\frac{\text { tempreature diffrences }}{\text { Thermal resistance }}\)
Temperature of the body, \(T_1=97^{\circ} \mathrm{F}=36.11^{\circ} \mathrm{C}\)
Temperature of the surroundings, \(T_2=47^{\circ} \mathrm{F}=8.33^{\circ} \mathrm{C}\) ‘
Conductivity of the cloth, \(\mathrm{K}=0.04 \mathrm{~J} \mathrm{~s}^{-1} \mathrm{~m}^{-1}{ }^{\circ} \mathrm{C}\)
Thickness of the cloth \(\mathrm{I}=0.5 \mathrm{~cm}=0.005 \mathrm{~m}\)
Area of the cloth, \(A=1.6 \mathrm{~m}^2\)
Difference in the temperature \(\Delta T=T_1-T_2=36.11-8.33=27.78^{\circ} \mathrm{C}\)
Thermal resistance \(=\frac{l}{K A}=\frac{0.005}{(0.04) \times 1.6}=0.078125\)
Rate at which heat is flowing out is given by
\(
\begin{aligned}
& \frac{\Delta Q}{\Delta t}=\frac{T_1-T_2}{\frac{l}{K A}} \\
& \frac{\Delta Q}{\Delta t}=\frac{27.78}{0.078125} \\
& \frac{\Delta Q}{\Delta t}=356 \mathrm{~J} / \mathrm{s}
\end{aligned}
\)

Hint

Area of the bottom of the container, \(A=25 \mathrm{~cm}^2=25 \times 10^{-4} \mathrm{~m}^2\)
Thickness of the vaporisation of water, \(l=1 \mathrm{~mm}=10^{-3} \mathrm{~m}\).
Latent heat of vaporisation of water, \(L=2.26 \times 106 \mathrm{J} \mathrm{Kg}^{-1}\)
Thermal conductivity of the container, \(\mathrm{K}=50 \mathrm{W} \mathrm{m}^{-1_{\circ}} \mathrm{C}^{-1}\) mass \(=100 \mathrm{~g}=0.1 \mathrm{Kg}\) mass \(=100 \mathrm{~g}=0.11 \mathrm{~kg}\)
Rate of heat transfer from the base of the container is given by
\(
\begin{aligned}
& \frac{\Delta Q}{\frac{\Delta}{t}}=\frac{m L}{\Delta t}=\frac{(0.1) \times 2.26 \times 10^6}{1 \mathrm{~min}} \\
& \frac{\Delta Q}{\frac{\Delta}{t}}=0.376 \times 10^4 \mathrm{~J} / \mathrm{s}
\end{aligned}
\)
Also,
\(
\begin{aligned}
& \frac{\Delta Q}{\frac{\Delta}{t}}=\frac{\Delta T}{\frac{l}{K A}} \\
& \Rightarrow 0.376 \times 10^4=\frac{T-100}{\frac{10^{-3}}{50 \times 25 \times 10^{-4}}} \\
& \Rightarrow 0.376 \times 10^4=\frac{50 \times 25 \times 10^{-4}(T-100)}{10^{-3}} \\
& \Rightarrow(\mathrm{T}-100)=3.008 \times 10 \\
& \Rightarrow \mathrm{T}=130^{\circ} \mathrm{C}
\end{aligned}
\)

Alternate:

\(
\begin{aligned}
& \frac{R_l}{K A}=\frac{1 \times 10^{-3}}{50 \times 25 \times 10^{-4}}=\frac{1}{125} \cdot{ }^{\circ} C / W \\
& i=\frac{\theta-100}{R}=125(\theta-100) \\
& i=m L=\frac{0.1}{60} \times 2.26 \times 10^6=3.8 \times 10^3 W \\
& 125(\theta-100)=3.8 \times 10^3 \\
& \theta=130^{\circ} C
\end{aligned}
\)

Hint

Given
\(
\begin{aligned}
& \mathrm{k}=45 \mathrm{J} \mathrm{s}^{-1} \mathrm{~m}^{-1_0} \mathrm{C}^{-1} \\
& 1=1 \mathrm{~m} \\
& \mathrm{~A}=0.04 \mathrm{~m}^2=4 \times 10^{-6} \mathrm{~m}^2
\end{aligned}
\)
We know that
\(
\begin{aligned}
& \frac{\mathrm{Q}}{\mathrm{T}}=\frac{\mathrm{kA}\left(\mathrm{T}_1-\mathrm{T}_2\right)}{1} \\
& =\frac{46 \times 4 \times 10^6 \times(100-0)}{1} \\
& =184 \times 10^{-4} \\
& \mathrm{~mL}=184 \times 10^{-4} \\
& \mathrm{~m}=\frac{184 \times 10^{-4}}{33.6 \times 10^5} \\
& \mathrm{~m}=5.5 \times 10^{-5} \mathrm{gm}
\end{aligned}
\)

Hint

Area of the walls of the box, \(A=2400 \mathrm{~cm}^2=2400 \times 10^{-4} \mathrm{~m}^2\)
Thickness of the ice box, \(l=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}\)
Thermal conductivity of the material of the box, \(K=0.06 \mathrm{~W} \mathrm{~m}^{-1}{ }^{\circ} \mathrm{C}^{-1}\)
Temperature of the water outside the box, \(T_1=20^{\circ} \mathrm{C}\)
Temperature of ice, \(T_2=0^{\circ} \mathrm{C}\)
Rate of flow of heat \(=\frac{\text { Temprature differences }}{\text { Thermal resistance }}\)
\(\Rightarrow \frac{\Delta Q}{\Delta t}=\frac{T_1-T_2}{\frac{l}{K A}}\)
\(\Rightarrow \frac{\Delta Q}{\Delta t}=\left(\frac{20}{2} \times 10^{-3}\right) \times 0.06 \times 2400 \times 10^{-4}\)
\(\Rightarrow \frac{\Delta Q}{\Delta t}=24 \times 6=144 \mathrm{~J} / \mathrm{s}\)
\(
\begin{aligned}
& \text { Rate at which the ice melts }=\frac{m L_f}{t} \\
& \Rightarrow \frac{\Delta Q}{\Delta t}=\left(\frac{m}{t}\right) L_f \\
& \Rightarrow 144=\left(\frac{m}{t}\right) \times 3.4 \times 10^5 \\
& \Rightarrow \frac{m}{t}=\frac{144}{3.4 \times 10^5} \mathrm{~kg} / \mathrm{s} \\
& \Rightarrow \frac{m}{t}=\frac{144 \times 60 \times 60}{3.4 \times 10^5} \mathrm{Kg} / \mathrm{s} \\
& \Rightarrow \frac{m}{t}=1.52 \mathrm{~kg} / \mathrm{h}
\end{aligned}
\)

Hint

Thickness of porous walls, \(I=1 \mathrm{~mm}=10-3 \mathrm{~m}\)
mass, \(m=10 \mathrm{~kg}\)
Latent heat of vapourisation, \(L v=2.27 \times 106 \mathrm{~J} / \mathrm{kg}\)
Thermal conductivity, \(\mathrm{K}=0.80 \mathrm{J} \mathrm{s}^{-1} \mathrm{~m}^{-1_{\circ}} \mathrm{C}^{-1}\)
\(
\Delta Q=2.27 \times 106 \times 10 \mathrm{~J}
\)
\(0.1 \mathrm{~g}\) of water evaporate in \(1 \mathrm{sec}\), so \(10 \mathrm{~kg}\) water will evaporate in \(105 \mathrm{~s}\)
\(
\begin{aligned}
& \Rightarrow \frac{\Delta Q}{\Delta t}=\frac{2.27 \times 107}{10^5} \\
& \Rightarrow \frac{\Delta Q}{\Delta t}=2.27 \times 10^2 \mathrm{~J} / \mathrm{s} \\
& \Rightarrow \frac{\Delta Q}{\Delta t}=\frac{\Delta T}{\frac{l}{k A}} \\
& \Rightarrow \frac{\Delta Q}{\Delta t}=\left(\frac{42-T}{10^{-3}}\right) \cdot 0.80 \times 2 \times 10^{-2} \\
& \Rightarrow 2.27 \times 10^2=\frac{42-\mathrm{T}}{10^{-3}} \times 0.80 \times 2 \times 10^{-2} \\
& \Rightarrow \mathrm{T}=27.8^{\circ} \mathrm{C} \\
& \Rightarrow \mathrm{T}=28^{\circ} \mathrm{C}
\end{aligned}
\)

Hint

Thermal conductivity, \(K=45 \mathrm{~W} \mathrm{~m}^{-1}{ }^{\circ} \mathrm{C}^{-1}\)
Length, \(l=60 \mathrm{~cm}=0.6 \mathrm{~m}\)
Area of cross section, \(A=0.2 \mathrm{~cm}^2=0.2 \times 10^{-4} \mathrm{~m}^2\)
Initial temperature, \(T_1=40^{\circ} \mathrm{C}\)
Final temperature, \(T_2=20^{\circ} \mathrm{C}\)
\(
\frac{\Delta Q}{\Delta t}=\frac{K A\left(T_1-T_2\right)}{l}
\)
\(\frac{\Delta Q}{\Delta t}=\frac{45 \times 0.2 \times 10^{-4}(40-20)}{0.6}\)
\(=0.03 W\)

Hint

Area of cross section, \(A=10 \mathrm{~cm}^2=10 \times 10^{-4} \mathrm{~m}^2\)
Thermal conductivity, \(K=200 \mathrm{Js}^{-1} \mathrm{~m}^{-1}{ }^{\circ} \mathrm{C}^{-1}\)
Height, \(H=10 \mathrm{~cm}\)
Length, \(l=1 \mathrm{~mm}=10^{-3} \mathrm{~m}\)
Temperature inside the cylindrical vessel, \(T_1=50^{\circ} \mathrm{C}\)
temperature outside the vessel, \(T_2=30^{\circ} \mathrm{C}\)
\(
\text { Rate of flow of heat from } 1 \text { flat surface will be given by }
\)
\(
\begin{aligned}
& \frac{\Delta Q}{\Delta t}=\frac{T_1-T_2}{\frac{l}{K A}} \\
& \frac{\Delta Q}{\Delta t}=\frac{(50-30) \times 200 \times 10^{-3}}{10^{-3}} \\
& \frac{\Delta Q}{\Delta t}=6000 \mathrm{~J} / \mathrm{s}
\end{aligned}
\)
Heat escapes out from both the flat surfaces.
Net rate of heat flow \(=2 \times 6000=12000 \mathrm{~J} / \mathrm{sec}\)
\(
\begin{aligned}
& \frac{\Delta Q}{\Delta t}=\frac{m . s . \Delta T}{\Delta t} \\
& \text { Mass }=\text { Volume density } \\
& \Rightarrow 10^{-3} \times 0.1 \times 1000=0.1 \mathrm{~g}
\end{aligned}
\)
Using this in the above formula for finding the rate of flow of heat, we get
\(
\begin{aligned}
& 12000=0.1 \times 4200 \times \frac{\Delta T}{t} \\
& \Rightarrow \frac{\Delta T}{\Delta t}=\frac{12000}{420}=28.57
\end{aligned}
\)
As \(\Delta T=1^{\circ} \mathrm{C}\)
\(
\begin{aligned}
& \Rightarrow \frac{1}{t}=28.57 \\
& \Rightarrow t=\frac{1}{28.57}=0.035 \mathrm{sec}
\end{aligned}
\)

Hint

Area of cross section, \(A=0.2 \mathrm{~cm}^2=0.2 \times 10^{-4} \mathrm{~m}^2\)
Thermal conductivity, \(k=385 \mathrm{~W} \mathrm{~m}^{-1}{ }^{\circ} \mathrm{C}^{-1}\)
Rate of flow of heat \(=\frac{\text { temperature diffrences }}{\text { Thermal resistance }}\)
\(
\begin{aligned}
& \frac{\Delta Q}{\Delta t}=\frac{K A\left(T_1-T_2\right)}{l} \\
& \frac{\Delta Q}{\Delta t}=\left(\frac{80-20}{0.2}\right) \times 385 \times 0.2 \times 10^{-4} \\
& =2310 \times 10^{-3} \\
& =2.31 \mathrm{~J} / \mathrm{sec}
\end{aligned}
\)
Let te temperature of point \(C\) be \(T\), which is at a distance of \(11 \mathrm{~cm}\) from the left end. Rate of flow of heat is given by
\(
\begin{aligned}
& \frac{\Delta Q}{\Delta t}=\frac{K A \Delta T}{l} \\
& \Rightarrow \frac{\Delta T}{l}=\left(\frac{\Delta Q}{\Delta t}\right) \times \frac{1}{K A} \\
& \frac{T-20}{11 \times 10^{-2}}=\frac{2.31}{383 \times 0.2 \times 10^{-4}} \\
& \Rightarrow \mathrm{T}=33+20 \\
& \Rightarrow \mathrm{T}=53^{\circ} \mathrm{c}
\end{aligned}
\)

Hint

No heat will flow when, the temp at that points also \(25^{\circ} \mathrm{C}\)

\(
\text { i.e. } Q_{A B}=Q_{B C}
\)

i.e. \(\mathrm{Q}_{\mathrm{AB}}=\mathrm{Q}_{\mathrm{AB}}\)
So, \(\frac{\mathrm{KA}(100-25)}{100-\mathrm{x}}=\frac{\mathrm{KA}(25-0)}{\mathrm{x}}\)
\(\Rightarrow 75 \mathrm{x}=2500-25 \mathrm{x} \Rightarrow 100 \mathrm{x}=2500 \Rightarrow \mathrm{x}=25 \mathrm{~cm}\) from the end with \(0^{\circ} \mathrm{C}\)

Hint

\(
\begin{aligned}
& V=216 \mathrm{~cm}^3 \\
& a=6 \mathrm{~cm}, \text { Surface area }=6 \mathrm{a}^2=6 \times 36 \mathrm{~m}^2 \\
& l=0.1 \mathrm{~cm} \\
& \mathrm{Q} / \mathrm{t}=100 \mathrm{~W}, \\
& \mathrm{Q} / \mathrm{t}=\left(\mathrm{KA}\left(\theta_1-\theta_2\right) / {l}\right. \\
& \Rightarrow 100=\left(\mathrm{k} \times 6 \times 36 \times 10^{-4} \times 5\right) /\left(0.1 \times 10^{-2}\right) \\
& \Rightarrow \mathrm{k}=100 /\left(6 \times 36 \times 5 \times 10^{-1}\right)=0.9259 \mathrm{~W} / \mathrm{m}^{\circ} \mathrm{C} \approx 0.92 \mathrm{~W} / \mathrm{m}^{\circ} \mathrm{C}
\end{aligned}
\)

Hint

Temperature of water, \(T_1=1^{\circ} \mathrm{C}\)
Temperature if ice bath, \(T_2=0^{\circ} \mathrm{C}\)
Thermal conductivity, \(K=0.5 \mathrm{~W} / \mathrm{m}^{\circ} \mathrm{C}\)
Length through which heat is lost, \(l=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}\)
Area of cross-section, \(A=5 \times 10^{-2} \mathrm{~m}^2\)
Velocity of the block, \(v=10 \mathrm{~cm} / \mathrm{sec}=0.1 \mathrm{~m} / \mathrm{s}\)
Let the mass of the block be \(m\).
Power \(=\mathrm{F} \cdot {v}\)
\(=(\mathrm{mg}) v \dots(1)\)
Also,
\(
\begin{aligned}
& \text { Power }=\frac{\Delta Q}{\Delta t} \ldots \dots (2) \\
& \frac{\Delta Q}{\Delta t}=\frac{k . A\left(T_1-T_2\right)}{l} \dots(3)
\end{aligned}
\)
From equation (1), (2) and (3), we get
\(
\begin{aligned}
& (m g) v=\frac{k . A\left(T_1-T_2\right)}{l} \\
& m=\frac{0.5 \times 5 \times 140^{-2}(1)}{\left(2 \times 10^{-3}\right) \times 10 \times 0.1} \\
& m=12.5 \mathrm{~kg}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \mathrm{K}=1.7 \mathrm{~W} \mathrm{~m}^{-1 \circ} \mathrm{C}^{-1}, \text { Density of water, } \rho_\omega=10^2 \mathrm{~kg} / \mathrm{m}^3 \\
& \mathrm{~L}_{\text {ice }}=3.36 \times 10^5 \mathrm{~J} / \mathrm{kgT}=10 \mathrm{~cm}=10 \times 10^{-2} \mathrm{~m}
\end{aligned}
\)
(a) \(\frac{Q}{t}=\frac{K A\left(\theta_1-\theta_2\right)}{\ell} \Rightarrow \frac{\ell}{\mathrm{t}}=\frac{\operatorname{KA}\left(\theta_1-\theta_2\right)}{Q}=\frac{K A\left(\theta_1-\theta_2\right)}{\mathrm{mL_{ice}}}\)
\(
\begin{aligned}
& =\frac{K A\left(\theta_1-\theta_2\right)}{A l \rho_\omega \mathrm{L_{ice}}}=\frac{1.7 \times[0-(-10)]}{10 \times 10^{-2} \times 1000 \times 3.36 \times 10^5} \\
& =\frac{17}{3.36} \times 10^{-7}=5.059 \times 10^{-7} \approx 5 \times 10^{-7} \mathrm{~m} / \mathrm{sec}
\end{aligned}
\)

(b) let us assume that \(x\) length of ice has become formed to form a small strip of ice of length \(d x, d t\) time is required.
\(
\begin{aligned}
& \frac{d Q}{d t}=\frac{K A(\Delta \theta)}{x} \Rightarrow \frac{d m L}{d t}=\frac{K A(\Delta \theta)}{x} \\
& \Rightarrow \frac{A d x f \omega L}{d t}=\frac{K A(\Delta \theta)}{x}
\end{aligned}
\)
\(
\Rightarrow \frac{d x f \omega L}{d t}=\frac{K(\Delta \theta)}{x} \Rightarrow d t=\frac{x d x f \omega L}{K(\Delta \theta)}
\)
\(
\Rightarrow \int_0^t d t=\frac{f \omega L}{K(\Delta \theta)} \int_0^t x d x \Rightarrow t=\frac{f \omega L}{K(\Delta \theta)}\left[\frac{x^2}{2}\right]_0^T=\frac{f \omega L}{K \Delta \theta} \frac{l}{2}
\)
Putting values
\(
\begin{aligned}
& \Rightarrow t=\frac{1000 \times 3.36 \times 10^5 \times\left(10 \times 10^{-2}\right)^2}{1.7 \times 10 \times 2} \\
& =\frac{3.36}{2 \times 17} \times 10^6 \mathrm{sec} .=\frac{3.36 \times 10^6}{2 \times 17 \times 3600} \mathrm{hrs}=27.45 \mathrm{hrs} \approx 27.5 \mathrm{hrs}
\end{aligned}
\)

Hint

Let the point upto which ice is formed is at a distance of \(\mathrm{x} \mathrm{m}\) from the top of the lake.
Under steady state, the rate of flow of heat from ice to this point should be equal to the rate flow of heat from water to this point.
Temperature of the top layer of ice \(=-10^{\circ} \mathrm{C}\)
Temperature of water at the bottom of the lake \(=4^{\circ} \mathrm{C}\)
Temperature at the point upto which ice is formed \(=\left(\frac{\Delta Q}{\Delta t}\right)_{\text {ice }}=\left(\frac{\Delta Q}{\Delta t}\right)_{\text {water }}\)
When heat current in water and ice is same, steady state is reached.
\(
\begin{aligned}
& \frac{k_{\text {ice }}}{\frac{x}{A \times 10}}=\frac{k_{\text {water }}}{\frac{1-x}{A \times 4}} \\
& \frac{1.7 \times 10}{x}=\frac{0.5 \times 4}{1-x} \\
& \frac{1.7}{x}=\frac{0.5 \times 4}{1-x} \\
& \Rightarrow x=\frac{17}{19}=0.89 \mathrm{~m} \\
& \Rightarrow x=89 \mathrm{~cm}
\end{aligned}
\)

Hint

Thermal conductivity of \(\operatorname{rod} A B, K_{A B}=50 \mathrm{~J} \mathrm{~s}^{-1} \mathrm{~m}^{-1_0} \mathrm{C}^{-1}\)
Temperature of junction at \(\theta_A=40^{\circ} \mathrm{C}\)
Thermal conductivity of rod BC, \(K_{\mathrm{BC}}=200 \mathrm{~J} \mathrm{~s}^{-1} \mathrm{~m}^{-1_0} \mathrm{C}^{-1}\)
Temperature of junction at \(\theta_{\mathrm{B}}=80^{\circ} \mathrm{C}\)
Thermal conductivity of rod \(\mathrm{BC}, K_{C A}=400 \mathrm{~J} \mathrm{~s}^{-1} \mathrm{~m}^{-1_0} \mathrm{C}^{-1}\)
temperature of junction at \(\theta_{\mathrm{C}}=80^{\circ} \mathrm{C}\)
Length \(=20 \mathrm{~cm}=20 \times 10^{-2} \mathrm{~m}\)
\(A=1 \mathrm{~cm}^2=1 \times 10^{-4} \mathrm{~m}^2\)
(a) \(\frac{Q_{A B}}{t}=\frac{K_{A B} \times A\left(\theta_B-\theta_A\right)}{l}=\frac{50 \times 1 \times 10^{-4} \times 40}{20 \times 10^{-2}}=1 \mathrm{~W}\).
(b) \(\frac{Q_{A C}}{t}=\frac{K_{A C} \times A\left(\theta_C-\theta_A\right)}{l}=\frac{400 \times 1 \times 10^{-4} \times 40}{20 \times 10^{-2}}=800 \times 10^{-2}=8 \mathrm{~W}\)
(c) \(\frac{Q_{B C}}{t}=\frac{K_{B C} \times A\left(\theta_B-\theta_C\right)}{l}=\frac{200 \times 1 \times 10^{-4} \times 0}{20 \times 10^{-2}}=0\)

Hint

Let \(A\) be the area of cross section and \(K\) be the thermal conductivity of the material of the rod.
Let \(q_1\) be the rate of flow of heat through a semicircular rod.
Rate of flow of heat is given by
\(
q 1=\frac{d Q}{d t}=\frac{K \cdot A\left(T_1-T_2\right)}{\pi r}
\)
Let \(q_2\) be the rate of flow of heat through a straight rod.
\(
q_2=\frac{d Q}{d t}=\frac{K A\left(T_1-T_2\right)}{2 r}
\)
Ratio of the rate of flow of heat through the 2 rods
\(
=\frac{q 1}{q 2}=\frac{2 r}{\pi r}=\frac{2}{\pi}
\)

Hint

Let the temperatures at the ends \(A\) and \(B\) be \(T_A\) and \(T_B\), respectively.
Rate of flow of heat at the end \(A\) of the rod is given by
\(
\frac{\mathrm{d} Q_{\mathrm{A}}}{\mathrm{d} t}=K \mathrm{~A} \cdot \frac{\mathrm{d}}{\mathrm{d} l}\left(T_{\mathrm{A}}\right)
\)
The rate of flow of heat at the end \(\mathrm{B}\) of the rod is given by
\(
\frac{\mathrm{d} Q_{\mathrm{B}}}{\mathrm{d} t}=K \mathrm{~A} \cdot \frac{\mathrm{d}}{\mathrm{d} l}\left(T_{\mathrm{B}}\right)
\)
Heat absorbed by the rod \(=m s \Delta T\)
Here, \(s\) is the specific heat of the rod, and \(\Delta T\) is the temperature difference between ends \(A\) and \(B\).
The rate of heat absorption by the rod is given by
\(
\begin{aligned}
& \frac{\mathrm{d} Q}{\mathrm{~d} t}=m s \frac{\mathrm{d} T}{\mathrm{~d} t} \therefore m s \frac{\mathrm{dT}}{\mathrm{d} t}=K A \cdot \frac{\mathrm{d} \mathrm{T}_{\mathrm{A}}}{\mathrm{d} l}-K \mathrm{~A} \cdot \frac{\mathrm{dT}_{\mathrm{B}}}{\mathrm{d} l} \\
& \Rightarrow(0.4) \cdot \frac{\mathrm{dT}}{\mathrm{d} t}=200 \times 1 \times 10^{-4}(5-2.5) \\
& \frac{\mathrm{dT}}{\mathrm{d} t}=12.5^{\circ} \mathrm{C} / \mathrm{sec}
\end{aligned}
\)

Hint

Given
\(
\begin{aligned}
& \mathrm{K}_{\text {rubber }}==0.15 \mathrm{~J} \mathrm{~s}^{-1} \mathrm{~m}^{-1_0} \mathrm{C}^{-1} \\
& \mathrm{T}_2-\mathrm{T}_1=90^{\circ} \mathrm{C}
\end{aligned}
\)
We know for radial conduction in a Cylinder
\(
\begin{aligned}
& \frac{Q}{t}=\frac{2 \pi \mathrm{Kl}\left(T_2-T_1\right)}{\ln \left(R_2 / R_1\right)} \\
& =\frac{2 \times 3.14 \times 15 \times 10^{-2} \times 50 \times 10^{-1} \times 90}{\ln (1.2 / 1)}=232.5 \approx 233 \mathrm{j} / \mathrm{s}
\end{aligned}
\)

Hint

\(\mathrm{dQ} / \mathrm{dt}=\) Rate of flow of heat
Let us consider a strip at a distance \(r\) from the center of thickness dr.
\(d Q / d t=(K \times 2 \pi r d \times d \theta) / d r[d \theta=\) Temperature diff across the thickness \(d r]\)
\(
\Rightarrow C=\frac{\mathrm{K} \times 2 \pi \mathrm{rd} \times \mathrm{d} \theta}{\mathrm{dr}} \quad\left[\mathrm{c}=\frac{\mathrm{d} \theta}{\mathrm{dr}}\right]
\)
\(
\Rightarrow \mathrm{C} \frac{\mathrm{dr}}{\mathrm{r}}=\mathrm{K} 2 \pi \mathrm{d} \mathrm{d} \theta
\)
Integrating
\(
\Rightarrow C \int_{r_1}^{r_2} \frac{d r}{r}=\mathrm{K} 2 \pi d \int_{\theta_1}^{\theta_2} d \theta
\)
\(
C\left(\log r_2-\log r_1\right)=K 2 \pi d\left(\theta_2-\theta_1\right)
\)
\(
C=\frac{\mathrm{K} 2 \pi \mathrm{d}\left(\theta_2-\theta_1\right)}{\log \left(\mathrm{r}_2 / \mathrm{r}_1\right)}
\)

Hint

When the flat ends are maintained at temperatures \(T_1\) and \(T_2\) (where \(T_2>T_1\) ):
Area of cross-section through which heat is flowing, \(=A=\pi\left(R_2^2-R_1^2\right)\)
\(
\text { So, } Q=\frac{K A\left(T_2-T_1\right)}{l}=\frac{K \pi\left(R_2^2-R_1^2\right)\left(T_2-T_1\right)}{l}
\)
Considering a concentric cylindrical shell of radius ‘ \(r\) ‘ and thickness ‘dr’. The radial heat flow through the shell
\(
H=\frac{d Q}{d t}=-K A \frac{d \theta}{d t} \quad[(-) \text { ve because as } r-\text { increases } \theta \text {decreases}.
\)
\(
\mathrm{A}=2 \pi r {l}
\)
\(\mathrm{H}=-2 \pi \mathrm{rl} \mathrm{K} \frac{\mathrm{d} \theta}{\mathrm{dt}}\)
\(
\text { or } \int_{\mathrm{R}_1}^{\mathrm{R}_2} \frac{\mathrm{dr}}{\mathrm{r}}=-\frac{2 \pi \mathrm{lK}}{\mathrm{H}} \int_{\mathrm{T}_1}^{\mathrm{T}_2} \mathrm{~d} \theta
\)
Integrating and simplifying we get
\(
H=\frac{d Q}{d t}=\frac{2 \pi K l\left(T_2-T_1\right)}{\operatorname{Log_e}\left(R_2 / R_1\right)}=\frac{2 \pi K l\left(T_2-T_1\right)}{\ln \left(R_2 / R_1\right)}
\)

Hint

s equivalent to the series combination of 2 resistors.
\(
\therefore R_{\mathrm{S}}=R_1+R_2
\)
Resistance of a conducting slab, \(R=\frac{l}{K A}\)
\(
\frac{L_1+L_2}{K_S A}=\frac{L_1}{K_1 A}+\frac{L_2}{K_2 A}
\)
\(
\begin{aligned}
& \frac{L_1+L_2}{K_s}=\frac{L_1}{K_1}+\frac{L_2}{K_2} \\
& \frac{L_1+L_2}{K_s}=\frac{L_1 K_2+L_2 K_1}{K_1 \times K_2} \\
& K_s=\frac{\left(L_1+L_2\right)\left(K_1 K_2\right)}{L_1 k_2+L_2 K_1}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \mathrm{K}_{\mathrm{Cu}}=390 \mathrm{~W} \mathrm{~m}^{-1 \circ} \mathrm{C}^{-1} \\
& \mathrm{K}_{\mathrm{St}}=46 \mathrm{~W} \mathrm{~m}^{-1 \circ} \mathrm{C}^{-1}
\end{aligned}
\)
Now, Since they are in series connection, So, the heat passed through the crosssections is the same.
\(
\begin{aligned}
& \text { So, } Q_1=Q_2 \\
& \text { Or } \frac{K_{\mathrm{Cu}} \times \mathrm{A} \times(\theta-0)}{\mathrm{l}}=\frac{\mathrm{K}_{\mathrm{St}} \times \mathrm{A} \times(100-\theta)}{\mathrm{l}} \\
& \Rightarrow 390(\theta-0)=46 \times 100-46 \theta \Rightarrow 436 \theta=4600 \\
& \Rightarrow \theta=\frac{4600}{436}=10.55 \approx 10.6^{\circ} \mathrm{C}
\end{aligned}
\)

Hint

As the Aluminum rod and Copper rod joined are in parallel
\(
\begin{aligned}
& \frac{Q}{t}=\left(\frac{Q}{t}\right)_{\text {Al }}+\left(\frac{Q}{t}\right)_{C u} \\
& \Rightarrow \frac{K A\left(\theta_1-\theta_2\right)}{l}=\frac{K_1 A\left(\theta_1-\theta_2\right)}{l}+\frac{K_2 A\left(\theta_1-\theta_2\right)}{l} \\
& \Rightarrow K=K_1+K_2=(390+200)=590 \\
& \frac{Q}{t}=\frac{K A\left(\theta_1-\theta_2\right)}{l}=\frac{590 \times 1 \times 10^{-4} \times(60-20)}{1} \\
& =590 \times 10^{-4} \times 40=2.36 \mathrm{J}
\end{aligned}
\)

Hint

Area of cross section, \(A=0.20 \mathrm{~cm}^2=0.2 \times 10^{-4} \mathrm{~m}^2\)
Thermal conductivity of aluminium, \(K_{A L}=200 \mathrm{W} \mathrm{m}^{-1_0} \mathrm{C}^{-1}\)
Thermal conductivity of copper, \(K_{C u}=400 \mathrm{W} \mathrm{m}^{-1_0} \mathrm{C}^{-1}\).
\(\mathrm{l}=20 \mathrm{~cm}=2 \times 10^{-1} \mathrm{~m}\)
\(\text { Heat drawn per second }\)
\(
\mathrm{Q}_{\mathrm{Al}}+\mathrm{Q}_{\mathrm{Cu}}
\)
\(
\begin{aligned}
& =\frac{K_{A I} \times A \times(80-40)}{l}+\frac{K_{c u} \times A \times(80-40)}{l} \\
& =\frac{200 \times 0.2 \times 10^{-4} \times 40}{0.2}+\frac{400 \times 0.2 \times 10^{-4} \times 40}{0.2} \\
& =8 \times 10^{-1}+16 \times 10^{-1} \\
& =24 \times 10^{-1} \\
& =2.4 \mathrm{~J} / \mathrm{s}
\end{aligned}
\)
Heat drawn in 1 minute \(=2.4 \times 60=144 \mathrm{~J}\)

Hint

\(
(\mathrm{Q} / \mathrm{t})_{\mathrm{AB}}=(\mathrm{Q} / \mathrm{t})_{\mathrm{BE} \text { bent }}+(\mathrm{Q} / \mathrm{t})_{\mathrm{BE}}
\)
\(
(\mathrm{Q} / \mathrm{t})_{\mathrm{BE} \text { bent }}=\frac{\mathrm{KA}\left(\theta_1-\theta_2\right)}{70}
\)
\(
(\mathrm{Q} / \mathrm{t})_{\mathrm{BE}}=\frac{\mathrm{KA}\left(\theta_1-\theta_2\right)}{60}
\)
\(
\frac{(\mathrm{Q} / \mathrm{t})_{\mathrm{BEbent}}}{(\mathrm{Q} / \mathrm{t})_{\mathrm{BE}}}=\frac{60}{70}=\frac{6}{7}
\)
\(
(\mathrm{Q} / \mathrm{t})_{\mathrm{BE} \text { bent }}+(\mathrm{Q} / \mathrm{t})_{\mathrm{BE}}=130
\)
\(
\Rightarrow(Q / t)_{B E} \text { bent }+(Q / t)_{B E} \times 7 / 6=130
\)
\(
\Rightarrow\left(\frac{7}{6}+1\right)(Q / t)_{\mathrm{BE} \text { bent }}=130
\)
\(
(\mathrm{Q} / \mathrm{t})_{\mathrm{BE} \text { bent }}=\frac{130 \times 6}{13}=60 J
\)

Alternate:

Length \(A D=60 \mathrm{~cm}\)
Length of bent part \(=5+60+5=70 \mathrm{~cm}\)
Resistance \(\propto\) length
\(
\begin{aligned}
& R_1=70 R, R_2=60 R \\
& i_1=\frac{R_2}{R_1+R_2} i=\frac{60 R}{70 R+60 R} \times 130 \\
& =60 \mathrm{~J}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \frac{\mathrm{Q}}{\mathrm{t}} \text { bent }=\frac{780 \times \mathrm{A} \times 100}{70} \\
& \frac{\mathrm{Q}}{\mathrm{t}} \mathrm{str}=\frac{390 \times \mathrm{A} \times 100}{60} \\
& \frac{(\mathrm{Q} / \mathrm{t}) \text { bent }}{(\mathrm{Q} / \mathrm{t}) \mathrm{str}}=\frac{780 \times \mathrm{A} \times 100}{70} \times \frac{60}{390 \times \mathrm{A} \times 100}=\frac{12}{7}
\end{aligned}
\)

Hint

(a)
\(
\begin{aligned}
& \frac{Q}{t}=\frac{K A\left(\theta_1-\theta_2\right)}{1} \\
& =\frac{1 \times 2(40-32)}{2 \times 10^{-3}} \\
& =8000 \mathrm{~J} / \mathrm{sec} .
\end{aligned}
\)
(b) Resistance of glass \(=\frac{1}{a k_g}+\frac{1}{a k_g}\)
\(
\begin{aligned}
& \text { Resistance of air }=\frac{1}{a k_a} \\
& \text { Net resistance }=\frac{1}{a k_g}+\frac{1}{a k_g}+\frac{1}{a k_g} \\
& \left(\frac{2}{k_g}+\frac{1}{k_a}\right) \\
& =\frac{1}{a}\left(\frac{2 k_a+k g}{k_g k_a}\right) \\
& =\frac{1 \times 10^{-3}}{2}\left(\frac{2 \times 0.025+}{0.025}\right) \\
& =\frac{1 \times 10^{-3} \times 1.05}{0.05} \\
& \frac{Q}{t}=\frac{\left(\theta_1-\theta_2\right)}{R} \\
& =385.9=381 \mathrm{~W} .
\end{aligned}
\)

Hint

\(
\text { Now; Q/t remains same in both cases }
\)
\(
\begin{aligned}
& \text { In Case I : } \frac{\mathrm{K}_{\mathrm{A}} \times \mathrm{A} \times(100-70)}{\ell}=\frac{\mathrm{K}_{\mathrm{B}} \times \mathrm{A} \times(70-0)}{\ell} \\
& \Rightarrow 30 \mathrm{~K}_{\mathrm{A}}=70 \mathrm{~K}_{\mathrm{B}} \\
& \text { In Case II : } \frac{\mathrm{K}_{\mathrm{B}} \times \mathrm{A} \times(100-\theta)}{\ell}=\frac{\mathrm{K}_{\mathrm{A}} \times \mathrm{A} \times(\theta-0)}{\ell} \\
& \Rightarrow 100 \mathrm{~K}_{\mathrm{B}}-\mathrm{K}_{\mathrm{B}} \theta=\mathrm{K}_{\mathrm{A}} \theta \\
& \Rightarrow 100 \mathrm{~K}_{\mathrm{B}}-\mathrm{K}_{\mathrm{B}} \theta=\frac{70}{30} \mathrm{~K}_{\mathrm{B}} \theta \\
& \Rightarrow 100=\frac{7}{3} \theta+\theta \quad \Rightarrow \theta=\frac{300}{10}=30^{\circ} \mathrm{C}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \theta_1-\theta_2=100 \\
& \mathrm{Q} / \mathrm{t}=\left(\theta_1-\theta_2\right) / \mathrm{R} \\
& \mathrm{R}=\mathrm{R}_1+\mathrm{R}_2+\mathrm{R}_3=1 / \mathrm{AK}_{\mathrm{Al}}+1 / \mathrm{AK}_{\mathrm{Cu}}+l / \mathrm{AK_{ \textrm {Al } }} \\
& =\frac{\ell}{\mathrm{A}}\left(\frac{2}{200}+\frac{1}{400}\right)=\frac{\ell}{\mathrm{A}}\left(\frac{4+1}{400}\right)=\frac{\ell}{\mathrm{A}} \frac{1}{80} \\
& \frac{\mathrm{Q}}{\mathrm{t}}=\frac{100}{(\ell / \mathrm{A})(1 / 80)} \Rightarrow 40=80 \times 100 \times \frac{\mathrm{A}}{\ell} \\
& \Rightarrow \frac{\mathrm{A}}{\ell}=\frac{1}{200}
\end{aligned}
\)
for (b)
\(
\begin{aligned}
& R=R 1+R 2=R 1+\left(R_{C u} R_{A l}\right) /\left(R_{C u}+R_{A l}\right) \\
& =\frac{l}{K_{Al} A}+\frac{\frac{l}{K_{cu} A} \times \frac{l}{K_{Al} A}}{\frac{l}{K_{c u} A }+\frac{l}{K_{A L} A}}
& =\frac{l}{A \cdot K_{Al}}+\frac{l}{A\left(K_{Al}+K_{cu}\right)}
\end{aligned}
\)
\(
\begin{aligned}
& =\frac{l}{A}\left(\frac{1}{200}+\frac{1}{200+400}\right) \\
& =\frac{l}{A}\left(\frac{1}{200}+\frac{1}{600}\right) \\
& =\frac{4}{600} \frac{l}{A}
\end{aligned}
\)
\(
\frac{Q}{t}=\frac{\theta_1-\theta_2}{R}=\frac{100}{(l / A)(4 / 600)}=\frac{100 \times 600}{4} \frac{A}{l}=\frac{100 \times 600}{4} \times \frac{1}{200}=75
\)
for (c)
\(
\begin{aligned}
& \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}=\frac{1}{\frac{l}{\mathrm{AK}_{\mathrm{Al}}}}+\frac{1}{\frac{l}{\mathrm{AK}_{\mathrm{Cu}}}}+\frac{1}{\frac{l}{\mathrm{AK}_{\mathrm{Al}}}} \\
& =\frac{A}{l}\left(K_{Al}+K_{Cu}+K_{Al}\right)=\frac{A}{l}(2 \times 200+400)=\frac{A}{l}(800) \\
& \Rightarrow R=\frac{l}{A} \times \frac{1}{800} \\
&
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow \frac{Q}{t}=\frac{\theta_1-\theta_2}{R}=\frac{100 \times 800 \times A}{l} \\
& =\frac{100 \times 800}{200}=400 \mathrm{~W}
\end{aligned}
\)

Hint

\(
\text { Let the temp. at B be } T
\)
\(
\begin{aligned}
& \frac{Q_A}{t}=\frac{Q_B}{t}+\frac{Q_C}{t} \\
& \Rightarrow \frac{T_1-T}{l}=\frac{T-T_3}{3 l / 2}+\frac{T-T_2}{3 l / 2} \\
& \Rightarrow-7 T=-3 T_1-2\left(T_2+T_3\right) \\
& \Rightarrow \frac{K A\left(T_1-T\right)}{l}=\frac{K A\left(T-T_3\right)}{l+(l / 2)}+\frac{K A\left(T-T_2\right)}{l+(l / 2)} \\
& \Rightarrow 3 T_1-3 T=4 T-2\left(T_2+T_3\right) \\
& \Rightarrow T=\frac{3 T_1+2\left(T_2+T_3\right)}{7}
\end{aligned}
\)

Hint

(a) The temp at the both ends of bar \(F\) is same
Rate of Heat flow to right \(=\) Rate of heat flow through left
\(
\Rightarrow(\mathrm{Q} / \mathrm{t})_{\mathrm{A}}+(\mathrm{Q} / \mathrm{t})_{\mathrm{C}}=(\mathrm{Q} / \mathrm{t})_{\mathrm{B}}+(\mathrm{Q} / \mathrm{t})_{\mathrm{D}}
\)
\(
\Rightarrow \frac{K_{\mathrm{A}}\left(\mathrm{T}_1-T\right) \mathrm{A}}{l}+\frac{\mathrm{K}_{\mathrm{C}}\left(\mathrm{T}_1-\mathrm{T}\right) \mathrm{A}}{l}=\frac{\mathrm{K}_{\mathrm{B}}\left(T-\mathrm{T}_2\right) \mathrm{A}}{l}+\frac{\mathrm{K}_{\mathrm{D}}\left(T-\mathrm{T}_2\right) \mathrm{A}}{l}
\)
\(
\Rightarrow 2 \mathrm{~K}_0\left(\mathrm{~T}_1-\mathrm{T}\right)=2 \times 2 \mathrm{~K}_0\left(\mathrm{~T}-\mathrm{T}_2\right)
\)
\(
\Rightarrow \mathrm{T}_1-\mathrm{T}=2 \mathrm{~T}-2 \mathrm{~T}_2
\)
\(
\Rightarrow \mathrm{T}=\frac{\mathrm{T}_1+2 \mathrm{~T}_2}{3}
\)
(b) To find the rate of flow of heat from the source (rod G), which maintains a temperature \(T_2\) is given by Rate of flow of heat, \(\mathrm{q}=\frac{\Delta T}{\text { Thermal resistance }}\) . First, we will find the effective thermal resistance of the circuit. From the diagram, we can see that it forms a balanced Wheatstone bridge. Also, as the ends of rod \(\mathrm{F}\) are maintained at the same temperature, no heat current flows through rod \(\mathrm{F}\). Hence, for simplification, we can remove this branch. From the diagram, we find that \(R_A\) and \(R_B\) are connected in series.
\(\therefore \mathrm{R}_{\mathrm{AB}}=\mathrm{R}_{\mathrm{A}}+\mathrm{R}_{\mathrm{B}}\)
\(R_C\) and \(R_D\) are also connected in series.
\(\therefore R_{C D}=R_C+R_D\)
Then, \(R_{A B}\) and \(R_{C D}\) are in parallel connection.
\(
\begin{aligned}
& R_A=\frac{l}{K_0 A} \\
& R_B=\frac{l}{2 K_0 A} \\
& R_C=\frac{l}{K_0 A} \\
& R_D=\frac{l}{2 K_0 A} \\
& R_{A B}=\frac{3 l}{2 k_0 A} \\
& R_{C D}=\frac{3 l}{2 K_0 A} \\
& R_{\text {eff }}=\frac{\frac{3 l}{2 K_0 A} \times \frac{3 l}{2 K_0 A}}{\frac{(3 l)}{2 K_0 A}+\frac{3 l}{2 K_0 A}} \\
& =\frac{3 l}{4 K_0 A} \\
& q=\frac{\Delta T}{\mathrm{R}_{eff}} \\
& =\frac{T_1-T_2}{\frac{3 l}{4 K_0 A}} \\
& \Rightarrow \frac{4 K_0 A\left(T_1-T_2\right)}{3 l}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \operatorname{Tan} \phi=\frac{r_2-r_1}{L}=\frac{\left(y-r_1\right)}{x} \\
& \Rightarrow x_2-x_1=y L-r_1 L \\
& \text { Differentiating wr to ‘ } x \text { ‘ } \\
& \Rightarrow r_2-r_1=\frac{L d y}{d x}-0 \\
& \Rightarrow \frac{d y}{d x}=\frac{r_2-r_1}{L} \Rightarrow d x=\frac{d y L}{\left(r_2-r_1\right)} \cdots(1) \\
& \text { Now } \frac{Q}{T}=\frac{K \pi y^2 d \theta}{d x} \Rightarrow \frac{Q d x}{T}=k \pi y^2 d \theta \\
& \Rightarrow \frac{Q L d y}{r_2 – r_1}=K \pi y^2 d \theta \\
& \Rightarrow d \theta =\frac{Q L d y}{\left(r_2-r_1\right) K \pi y^2}
\end{aligned}
\)
Integrating both side
\(
\begin{aligned}
& \Rightarrow \int_{\theta_1}^{\theta_2} d \theta=\frac{Q L}{\left(r_2-r_1\right) K \pi} \int_{r_1}^{r_2} \frac{d y}{y} \\
& \Rightarrow\left(\theta_2-\theta_1\right)=\frac{Q L}{\left(r_2-r_1\right) K \pi} \times\left[\frac{-1}{y}\right]_{r_1}^{r_2} \\
& \Rightarrow\left(\theta_2-\theta_1\right)=\frac{Q L}{\left(r_2-r_1\right) K \pi} \times\left[\frac{1}{r_1}-\frac{1}{r_2}\right] \\
& \Rightarrow\left(\theta_2-\theta_1\right)=\frac{Q L}{\left(r_2-r_1\right) K \pi} \times\left[\frac{r_2-r_1}{r_1+r_2}\right] \\
& \Rightarrow Q=\frac{K \pi r_1 r_2\left(\theta_2-\theta_1\right)}{L}
\end{aligned}
\)

Hint

Given:
Length of the rod, \(l=20 \mathrm{~cm}=0.2 \mathrm{~m}\)
Area of cross section of the rod, \(A=1.0 \mathrm{~cm}^2=1.0 \times 10^{-4} \mathrm{~m}^2\)
Thermal conductivity of the material of the rod, \(\mathrm{k}=200 \mathrm{~W} \mathrm{~m}^{-1_{\circ}} \mathrm{C}^{-1}\)
The temperature of one end of the rod is increased uniformly by \(60^{\circ} \mathrm{C}\) within 10 minutes.
This mean that the rate of increase of the temperature of one end is \(0.1^{\circ} \mathrm{C}\) per second
\(
\Rightarrow \frac{60}{10 \times 60} { }^{\circ} \mathrm{C} / \mathrm{s}
\)
So, total heat flow can be found by adding heat flow every second.
Rate of flow of heat \(=\frac{d Q}{d t}\)
\(
\mathrm{Q}_{\text {net }}=\Sigma \frac{K A}{d}\left(T_2-T_1\right) \times \Delta t
\)
For each interval,
\(
\begin{aligned}
& \Delta t=1 \\
&\mathrm{Q}_{\text {net }}= \frac{\mathrm{KA} \times 0.1}{\mathrm{~d}}+\frac{\mathrm{KA} \times 0.2}{\mathrm{~d}}+\ldots \ldots .+\frac{\mathrm{KA} \times 60}{\mathrm{~d}} \\
& \mathrm{Q}_{\text {net }}=\frac{K A}{d}(0.1+0.2+\ldots \ldots \ldots+60.0)
\end{aligned}
\)
\(
[\therefore a+2 a+\ldots \ldots \ldots+n a=n / 2\{2 a+(n-1) a\}]
\)
\(
\mathrm{Q}_{\text {net }}=\frac{K A}{d} \times \frac{600}{2}(0.1+60)
\)
\(
=\frac{200 \times-10^{-4}}{20 \times 10^{-2}} \times \frac{600}{2} \times 60.1
\)
\(
\mathrm{Q}_{\text {net }}=1800 \mathrm{~J} \text { (approximately) }
\)

Hint

\(
\begin{aligned}
& a=r_1=5 \mathrm{~cm}=0.05 \mathrm{~m} \\
& b=r_2=20 \mathrm{~cm}=0.2 \mathrm{~m} \\
& \theta_1=T_1=50^{\circ} \mathrm{C}, \theta_2=T_2=10^{\circ} \mathrm{C}
\end{aligned}
\)
Now, considering a small strip of thickness ‘dr’ at a distance ‘ \(r\) ‘.

\(
\begin{aligned}
& A=4 \pi r^2 \\
& H=-4 \pi r^2 K \frac{d \theta}{d r}[(-) \text { ve because with increase of } r, \theta \text { decreases }] \\
& =\int_a^b \frac{d r}{r^2}=\frac{-4 \pi K}{H} \int_{\theta_1}^{\theta_2} d \theta \quad \text { On integration, } \\
& H=\frac{d Q}{d t}=K \frac{4 \pi a b\left(\theta_1-\theta_2\right)}{(b-a)}
\end{aligned}
\)
Putting the values we get
\(
\begin{aligned}
& \frac{\mathrm{K} \times 4 \times 3.14 \times 5 \times 20 \times 40 \times 10^{-3}}{15 \times 10^{-2}}=100 \\
& \Rightarrow \mathrm{K}=\frac{15}{4 \times 3.14 \times 4 \times 10^{-1}}=2.985 \approx 3 \mathrm{w} / \mathrm{m}{ }^{\circ} \mathrm{C}
\end{aligned}
\)

Hint

Rate of transfer of heat from the rod is given as
\(
\frac{\Delta Q}{\Delta t}=\frac{K A\left(T_1-T_2\right)}{l}
\)
In time t, the temperature difference becomes half.
In time \(\Delta t\), the heat transfer from the rod will be given by
\(
\Delta Q=\frac{K A\left(T_1-T_2\right)}{l} \Delta t \dots(i)
\)
Heat loss by water at temperature \(T_1\) is equal to the heat gain by water at temperature \(T_2\).
Therefore, heat loss by water at temperature \(T_1\) in time \(\Delta t\) is given by
\(
\Delta Q=m s\left(T_1-T_1 \prime\right) \dots(ii)
\)
From equations (i) and (ii),
\(
\mathrm{ms}\left(T_1-T_1 \prime\right)=\frac{K A\left(T_1-T_2\right)}{l} \Delta t
\)
\(\Rightarrow T_1 \prime=T_1-\frac{K A\left(T_1-T_2\right)}{l(m s) \Delta t}\)
This gives us the fall in the temperature of water at temperature \(T_1\).
Similarly, rise in temperature of water at temperature \(T_2\) is given by
\(
T_2 \prime=T_2+\frac{K A\left(T_1-T_2\right)}{l(m s)} \Delta t
\)
Change in the temperature is given by
\(
\begin{aligned}
& \left(T_1 \prime-T_2 \prime\right)=\left(T_1-T_2\right)-2 \frac{K A\left(T_1-T_2\right)}{l(m s)} \Delta t \\
& \Rightarrow\left(T_1-T_2\right)-\left(T_1 \prime-T_2 \prime\right)=2 \frac{K A\left(T_1-T_2\right)}{l(m s)} \Delta t \\
& \Rightarrow \frac{\Delta T}{\Delta t}=2 \frac{K A\left(T_1-T_2\right)}{l(m s)}
\end{aligned}
\)
Here, \(\frac{\Delta T}{\Delta t}\) is the rate of change of temperature difference.
Taking limit \(\Delta t \rightarrow 0\),
\(
\begin{aligned}
& \Rightarrow \frac{d T}{d t}=2 \frac{K A\left(T_1-T_2\right)}{l(m s)} \\
& \Rightarrow \frac{d T}{T_1-T_2}=2 \frac{K A}{l(m s) d t}
\end{aligned}
\)
On integrating within proper limit, we get
\(
\begin{aligned}
& \int_{\left(T_1-T_2\right)}^{\left(T_1-T_2\right) / 2} \frac{d T}{\left(T_1-T_2\right)}=2 \frac{K A}{l(m s)} \int_0^t d t \\
& \Rightarrow \operatorname{In}\left[\frac{\left(T_1-T_2\right)}{2\left(T_1-t_2\right)}\right]=2=\frac{K A}{l(m s)} t \\
& \Rightarrow \operatorname{In}[2]=2 \frac{K A}{l(m s)} t \\
& \Rightarrow t=\operatorname{In}[2] \frac{l m s}{2 K A}
\end{aligned}
\)

Hint

Rate of transfer of heat from the rod is given by
\(
\frac{\Delta Q}{\Delta t}=\frac{K A\left(T_2-T_1\right)}{l}
\)
Heat transfer from the rod in time \(\Delta t\) is given by
\(
\frac{\Delta Q}{\Delta t}=\frac{K A\left(T_2-T_1\right)}{l} \Delta t \dots(1)
\)
Heat loss by the body at temperature \(T_2\) is equal to the heat gain by the body at temperature \(T_1\)
Therefore, heat loss by the body at temperature \(T_2\) in time \(\Delta t\) is given by
\(\Delta Q=m_2 s_2\left(T_2-T_2^{\prime}\right) \ldots(2)\)
From equations (1) and (2)
\(
\begin{aligned}
& m_2 s_2\left(T_2-T_2 \prime\right)=\frac{K A\left(T_2-T_1\right)}{l} \Delta t \\
& \Rightarrow T_2 \prime=T_2-\frac{K A\left(T_2-T_1\right)}{l\left(m_2 s_2\right)} \Delta t
\end{aligned}
\)
This gives us the fall in the temperature of the body at temperature \(T_2\).
Similarly, rise in temperature of water at temperature \(T_1\) is given by
\(T_{1^{\prime}}=T_1+\frac{K A\left(T_2-T_1\right)}{l\left(m_1 s_1\right)} \Delta t\)
Change in the temperature is given by
\(
\begin{aligned}
& \left(T_{2^{\prime}}-T_1 \prime\right)=\left(T_2-T_1\right)-\left[\frac{K A\left(T_2-T_1\right)}{l m_1 s_1} \Delta t+\frac{K A\left(T_2-T_1\right)}{l m_2 s_1} \Delta t\right] \\
& \Rightarrow\left(T_2 \prime-T_1 \prime\right)-\left(T_2-T_1\right)=-\left[\frac{K A\left(T_2-T_1\right)}{l m_1 s_1} \Delta t+\left[\frac{K A\left(T_2-T_1\right)}{l m_2 s_2} \Delta t\right]\right. \\
& \Rightarrow \frac{\Delta T}{\Delta t}=\frac{K A\left(T_2-T_1\right)}{l}\left[\frac{1}{m_1 s_1}+\frac{1}{m_2 s_2}\right] \Delta t \\
& \Rightarrow \frac{1}{T_2-T_1} \Delta T=-\frac{K A}{l}\left[\frac{m_1 s_1+m_2 s_2}{m_1 s_1 m_2 s_2}\right]
\end{aligned}
\)
By integrating both sides, we get
\(\lim \Delta t \rightarrow 0\)
\(
\begin{aligned}
& \int \frac{1}{T_2-T_1} d T=\int-\frac{K A}{l}\left[\frac{m_1 s_1+m_2 s_2}{m_1 s_1 m_2 s_2}\right] d t \\
& \Rightarrow \operatorname{In}\left[T_2-T_1\right]=-\frac{K A}{l}\left[\frac{m_1 s_1+m_2 s_2}{m_1 s_1 m_2 s_2}\right] t \\
& \Rightarrow\left(T_2-T_1\right)=e^{-\lambda t}
\end{aligned}
\)
Here, \(\lambda=\mathrm{KA} / {l}\left[m_1 s_1+m_2 s_2 / m_1 s_1 m_2 s_2\right]\)

Hint

In time \(d t\), heat transfer through the bottom of the cylinder is given by \(\frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\mathrm{KA}\left(T_s-\mathrm{T}_0\right)}{x}\)
For a monoatomic gas, pressure remains constant.
\(
\begin{aligned}
& \therefore d Q=n C_p d T \\
& \therefore \frac{n C_p d T}{d t}=\frac{\mathrm{KA}\left(\mathrm{T}_2-\mathrm{T}_0\right)}{x}
\end{aligned}
\)
For a monoatomic gas,
\(
C_p=\frac{5}{2} R
\)
\(
\begin{aligned}
& \Rightarrow \frac{\mathrm{n} 5 \mathrm{RdT}}{2 \mathrm{dt}}=K A \frac{T_s-T_0}{x} \\
& \Rightarrow \frac{5 \mathrm{nR}}{2} \frac{\mathrm{dT}}{d t}=\frac{K A\left(T_s-T_0\right)}{x} \\
& \Rightarrow \frac{\mathrm{dT}}{T_s-T_0}=\frac{-2 \mathrm{KAdt}}{5 \mathrm{nRx}}
\end{aligned}
\)
Integrating both the sides with limit \(T_0\) to \(T\),
\(
\begin{aligned}
& \Rightarrow \operatorname{ln}\left(\frac{T_s-T}{T_s-T_0}\right)=\frac{-2 \mathrm{KAt}}{5 \mathrm{nRx}}
\end{aligned}
\)
\(
T_s-T=\left(T_s-T_0\right) e^{\frac{-2 \text { KAt }}{\text { 5nRx }}}
\)
\(
T-T_0=\left(T_s-T_0\right)\left[1-e^{\frac{-2 {KAt}}{5nRx}}\right]
\)
From the gas equation,
\(
\begin{aligned}
& \frac{P_a A l}{n R}=T-T_0 \\
& \therefore \frac{P_a A l}{n R}=\left(T_s-T_0\right)\left[1-e^{\frac{-2 K a t}{5 n R x}}\right] \\
& \Rightarrow l=\frac{n R}{P_a A}\left(T_s-T_0\right)\left[1-e^{\frac{-2 K A t}{5 n R x}}\right]
\end{aligned}
\)

Hint

Given:
Area of the body, \(A=1.6 \mathrm{~m}^2\)
Temperature of the body, \(T=310 \mathrm{~K}\)
From Stefan-Boltzmann law,

\(\text { { Energy radiated }/{Time} }=\sigma \mathrm{AT}^4\)
Here, \(A\) is the area of the body, and \(\sigma\) is the Stefan-Boltzmann constant.
Energy radiated per second \(=1.6 \times 6 \times 10^{-8} \times(310)^4\)
\(=886.58 \approx 887 \mathrm{~J}\)

Hint

Given:
Area of the body, \(A=12 \times 10^{-4} \mathrm{~m}^2\) Temperature of the body, \(T=(273+20)=293 \mathrm{~K}\) Emissivity, \(e=0.80\)
Rate of emission of heat, \(R=A e \sigma T^4\)
\(
\begin{aligned}
& R=12 \times 10^{-4} \times 0.80 \times 6.0 \times 10^{-8} \times(293)^4 \\
& R=0.42 \text { J (approximately) }
\end{aligned}
\)

Hint

\(C\)-cooling rate \(H\) – heat loss
\(
\frac{H_{A l}}{H_{C u}}=\frac{\sigma A_{A I}\left(T_{A I^4-T_0^4}\right)}{\sigma e A_{C u}\left(T_{C u^4-T_0^4}\right)}
\)
\(
\text { at } t=0 T_{A l}=T_{C u}=T=\frac{A_{A l}}{A_{C u}}=\frac{R^2}{(2 R)^2}=\frac{1}{4}
\)
(b) \(C-\) cooling rate
\(
\begin{aligned}
& \frac{C_{A l}}{C_{C u}}=\left(\frac{A_{A l}}{m_{A l} S_{A l}}\right)\left(\frac{m_{C u} S_{C u}}{A_{C u}}\right) \\
& =\left(\frac{A_{A l}}{A_{C u}}\right)\left(\frac{m_{C u}}{m_{A l}}\right)\left(\frac{S_{C u}}{S_{A l}}\right) \\
& =\left(\frac{R}{2 R}\right)^2 \times\left\{\left(\frac{2 R}{R}\right)^3 \times 3.4\right\} \times \frac{390}{900}=2.9:1
\end{aligned}
\)

Hint

Given:
Power of the bulb, \(P=100 \mathrm{~W}\)
Length of the filament, \(l=1.0 \mathrm{~m}\)
Radius of the filament, \(r=4 \times 10^{-5} \mathrm{~m}\)
According to Stefan’s law,
\(\left(\frac{\text { Energy radiated }}{\text { time }}\right)=e \cdot A \sigma T^4\)
Here, \(e\) is the emissivity of the tungsten and \(\sigma\) is Stefan’s constant.
\(
\begin{aligned}
& \therefore P=e \cdot\left(\pi r^2\right) \sigma T^4 \\
& \Rightarrow 100=(0.8) \times 3.14 \times\left(4 \times 10^{-5}\right) \times 6 \times 10^{-8} \times T^4 \\
& \Rightarrow T=1700 K \text { (approximately) }
\end{aligned}
\)

Hint

(a)
Area of the ball, \(A=20 \times 10^{-4} \mathrm{~m}^2\)
Temperature of the ball, \(T=57^{\circ} \mathrm{C}=57+273=330 \mathrm{~K}\)
Amount of heat radiated per second \(=A \sigma T^4\)
\(
\begin{aligned}
& =20 \times 10^{-4} \times 6 \times 10^{-8} \times(330)^4 \\
& =1.42 \mathrm{~J}
\end{aligned}
\)
(b)
Net rate of heat flow from the ball when its
emperature is \(200^{\circ} \mathrm{C}\) is given by
\(
\begin{aligned}
& \mathrm{eA} \sigma\left(T_1^4-T_2^4\right) \\
& =20 \times 10^{-4} \times 6 \times 10^{-8} \times 1\left((473)^4-(330)^4[\therefore e=1]\right. \\
& =4.58 \mathrm{~W}
\end{aligned}
\)

Hint

Given:
Radius of the spherical tungsten, \(r=10^{-2} \mathrm{~m}\)
Emissivity of the tungsten, \(e=0.3\)
Stefan’s constant, \(\sigma=6.0 \times 10^{-8} \mathrm{~W} \mathrm{~m}^{-2} \mathrm{~K}^{-4}\)
The surface area of the spherical tungsten, \(A=4 \pi r^2\)
\(
\begin{aligned}
& A=4 \pi\left(10^{-2}\right)^2 \\
& A=4 \pi \times 10^{-4} \mathrm{~m}^2
\end{aligned}
\)
Rate at which energy must be supplied is given by
\(
\begin{aligned}
& \operatorname{e\sigma A}\left(T_2^4-T_1^4\right) \\
& =(0.3) \times 6 \times 10^{-8} \times 4 \pi \times 10^{-4} \times\left((1000)^4-(300)^4\right) \\
& =22.42 \approx \text { watt }=22 \text { watt }
\end{aligned}
\)

Hint

It is given that a cube behaves like a black body.
\(\therefore\) Emissivity, \(e=1\)
Stefan’s constant, \(\sigma=6 \times 10^{-8} \mathrm{~W} /\left(\mathrm{m}^2 \mathrm{~K}^4\right)\)
Surface area of the cube, \(A=6 \times 25 \times 10^{-4}\)
Mass of the cube, \(m=1 \mathrm{~kg}\)
Specific heat capacity of the material of the cube, \(s=400 \mathrm{J} \mathrm{kg}^{-1} \mathrm{~K}^{-1}\)
Temperature of the cube, \(T_1=227+273=500 \mathrm{~K}\)
Temperature of the surrounding, \(T_0=27+273=300 \mathrm{~K}\)
Rate of flow of heat is given by
\(
\begin{aligned}
& \frac{\Delta Q}{\Delta}=\mathrm{eA \sigma}\left(\mathrm{T}^4-\mathrm{T}_0{ }^4\right) \\
& \Rightarrow \mathrm{ms} \cdot \frac{\Delta T}{\Delta t}=1 \times 6 \times 10 \times 25 \times 10^{-4}\left(500^4-300^4\right) \\
& \Rightarrow \frac{\Delta T}{\Delta t}=\frac{36 \times 25 \times\left(500^4-300^4\right) \times 10^{-12}}{400} \\
& \Rightarrow \frac{\Delta T}{\Delta t}=0.12^{\circ} \frac{C}{s}
\end{aligned}
\)

Hint

According to Stefan,s law,
power \(=e A \sigma\left(T^4-T_0^4\right)\)
Temperature difference \(=200 \mathrm{~K}\)
Let the emissivity of copper be \(e\).
\(
210=e A \sigma\left(500^4-300^4\right) \dots(1)
\)
When the surface of the copper sphere is completely blackened, \(700 \mathrm{~W}\) is needed to maintain the temperature of the sphere.
For a black body,
\(
\begin{aligned}
& e=1 \\
& 700=1 \cdot A \sigma(5004-3004) \dots(2)
\end{aligned}
\)
On dividing equation (1) by equation (2) we have,
\(
\begin{aligned}
& \frac{200}{700}=\frac{e}{1} \\
& \Rightarrow e=0.3
\end{aligned}
\)

Hint

Surface area of the spherical ball, \(S_A=20 \mathrm{~cm}^2=20 \times 10^{-4} \mathrm{~m}^{-2}\)
Surface area of the spherical shell, \(S_B=80 \mathrm{~cm}^2=80 \times 10^{-4} \mathrm{~m} 2\)
Temperature of the spherical ball, \(T_{\mathrm{A}}=300 \mathrm{~K}\)
Temperature of the spherical shell, \(T_{\mathrm{B}}=300 \mathrm{~K}\)
Radiation energy emitted per second by the spherical ball A is given by
\(
\begin{aligned}
& \mathrm{E}_{\mathrm{A}}=\sigma \mathrm{S}_{\mathrm{A}} \mathrm{T}_{\mathrm{A}}{ }^4 \\
& \Rightarrow \mathrm{E}_{\mathrm{A}} 6.0 \times 10^{-8} \times 20 \times 10^{-4} \times(300)^4 \\
& \Rightarrow \mathrm{E}_{\mathrm{A}}=0.97 \mathrm{~J}
\end{aligned}
\)
Radiation energy emitted per second by the inner surface of the spherical shell B is given by
\(
\begin{aligned}
& E_B=\sigma S_B T_B^4 \\
& \Rightarrow E_B=6.0 \times 10^{-8} \times 80 \times 10^{-4} \times(300)^4 \\
& \Rightarrow E_B=3.76 \mathrm{~J} \approx 3.8 \mathrm{~J}
\end{aligned}
\)
Energy emitted by the inner surface of \(B\) that falls back on its surface is given by
\(
\begin{aligned}
E & =E_B-E_A=3.76-0.94 \\
& \Rightarrow E=2.82 \mathrm{~J}
\end{aligned}
\)

Hint

\(
\mathrm{Q} / \mathrm{t}=\mathrm{eA \sigma}\left(\mathrm{T}_2^4-\mathrm{T}_1^4\right)
\)
\(
\begin{aligned}
& \Rightarrow \text { Q/At }=1 \times 6 \times 10^{-8}\left[(300)^4-(290)^4\right] \\
& =6 \times 10^{-8}\left(81 \times 10^8-70.7 \times 10^8\right)=6 \times 10.3
\end{aligned}
\)
\(
\begin{aligned}
& \frac{Q}{t}=\frac{K A\left(\theta_1-\theta_2\right)}{l} \\
& \Rightarrow \frac{Q}{t A}=\frac{K\left(\theta_1-\theta_2\right)}{l}=\frac{K \times 17}{0.5}=6 \times 10.3 \\
& =\frac{K \times 17}{0.5} \Rightarrow K=\frac{6 \times 10.3 \times 0.5}{17}=1.8 \mathrm{~W} \mathrm{~m}^{-1} { }^{\circ} \mathrm{C}^{-1}
\end{aligned}
\)

Hint

\(
\text { Heat flowing through the rod per second in steady state, }
\)
\(
\frac{d Q}{d t}=\frac{K A d \theta}{l} \dots(i)
\)
Heat radiated from the open end of the rod per second in steady state,
\(
\frac{d Q}{d t}=A \sigma\left(T^4-T_0^4\right) \dots(ii)
\)
From Eqs. (i) and (ii),
\(
\begin{aligned}
\frac{K d \theta}{l} & =\sigma\left(T^4-T_0^4\right) \\
\frac{K \times 50}{0.2} & =6.0 \times 10^{-8}\left[(7.5)^4-(3)^4\right] \times 10^8 \\
K & =74 \mathrm{W} \mathrm{m}^{-1} \mathrm{~K}^{-1}
\end{aligned}
\)

Hint

Given:
Volume of water, \(V=100 \mathrm{cc}=100 \times 10^{-3} \mathrm{~m}^3\)
Change in the temperature of the liquid, \(\Delta \theta=5^{\circ} \mathrm{C}\)
Time, \(T=5 \mathrm{~min}\)
For water,
\(
\begin{aligned}
& \frac{m s \Delta \theta}{t}=\frac{K A}{l}\left(T_1-T_0\right) \\
& \Rightarrow \frac{m s}{t}=\frac{K A}{l} \frac{\left(T_1-T_0\right)}{\Delta \theta} \\
& \Rightarrow \frac{100 \times 10^{-3} \times 1000 \times 4200}{5}=\frac{K A}{l} \frac{\left(313-T_0\right)}{\Delta \theta} \dots(i)
\end{aligned}
\)
For K-oil,
\(
\begin{aligned}
& \frac{m s}{t}=\frac{K A}{l} \frac{9\left(T_1-T_0\right)}{\Delta \theta} \\
& \Rightarrow \frac{m s}{t}=\frac{K A}{l} \frac{\left(T_1-T_0\right)}{\Delta \theta} \\
& \Rightarrow \frac{V p s}{t}=\frac{K A}{l} \frac{T_1-T_0}{\Delta \theta} \\
& \Rightarrow \frac{100 \times 10^{-3} \times 800 \times 2100}{t}=\frac{K A}{l} \frac{\left(313-T_0\right)}{\Delta \theta} \dots(ii)
\end{aligned}
\)
From (i) and (ii),
\(
\begin{aligned}
& \frac{100 \times 10^{-} 3 \times 800 \times 2100}{t}=\frac{100 \times 10^{-5} \times 1000 \times 4200}{5} \\
& \Rightarrow \mathrm{t}=\frac{5 \times 800 \times 2100}{1000 \times 4200}=\frac{2000}{1000} \\
& \Rightarrow \mathrm{t}=2 \mathrm{~min}
\end{aligned}
\)

Hint

Case-I: Initial temperature of the body \(=50^{\circ} \mathrm{C}\)
Final temperature of the body \(=45^{\circ} \mathrm{C}\)
Average temperature \(=47.5^{\circ} \mathrm{C}\)
Difference in the temperatures of the body and its surrounding \(=(47.5-\mathrm{T})^{\circ} \mathrm{C}\)
Rate of fall of temperature \(=\frac{\Delta t}{t}=\frac{5}{5}=1^{\circ} \mathrm{C} / \mathrm{min}\)
By Newton’s law of cooling,
\(
\frac{d T}{d t}=-K\left[T_{\text {avg }}-T_0\right]
\)
\(1=-K\left[47.5-T_0\right] \ldots \ldots \ldots . .(\mathrm{i})\)

Case-II: Initial temperature of the body \(=45^{\circ} \mathrm{C}\)
Final temperature of the body \(=40^{\circ} \mathrm{C}\)
Average temperature \(=42.5^{\circ} \mathrm{C}\)
Difference in the temperatures of the body and its surrounding \(=\left(42.5-\mathrm{T}_0\right)^{\circ} \mathrm{C}\)
Rate of fall of temperature \(=\frac{\Delta T}{t}=\frac{5}{8}=\frac{5}{8}^{\circ} \mathrm{C} / \mathrm{min}\)
From Newton’s law of cooling,
\(
\begin{aligned}
\frac{d T}{d t} & =-K\left[T_{a v g}-T_0\right] \\
0.625 & =-\mathrm{K}\left[42.5-\mathrm{T}_0\right] \dots(ii)
\end{aligned}
\)
Dividing (i) by (ii),
\(
\begin{aligned}
& \frac{1}{0.625}=\frac{47.5-T_0}{42.5-T_0} \\
& \Rightarrow 42.5-\mathrm{T}_{-} 0=29.68-0.625 \mathrm{~T} \\
& \Rightarrow \mathrm{T}_0=34^{\circ} \mathrm{C}
\end{aligned}
\)

Hint

Let water equivalent to calorimeter be \(w\).
Change in temperature \(=5^{\circ} \mathrm{C}\)
Specific heat of water \(=4200 \mathrm{~J} / \mathrm{Kg}^{\circ} \mathrm{C}\)
Rate of flow of heat is given by
\(q=\) Energy per unit time \(=\frac{m s \Delta T}{t}\)
Case 1:
\(q_1=\frac{\left(w+50 \times 10^{-3}\right) \times 4200 \times 5}{10}\)
Case 2
\(
q_2=\frac{\left(w+100 \times 10^{-3}\right) \times 4200 \times 5}{18}
\)
From calorimeter theory, these two rates of flow of heat should be equal to each other.
\(
\begin{aligned}
& \Rightarrow \mathrm{q}_1=\mathrm{q}_2 \\
& \frac{\left(w+50 \times 10^{-3}\right) \times 4200 \times 5}{10}=\frac{\left(w+100 \times 10^{-3}\right) \times 4200 \times 5}{18} \\
& \Rightarrow 18\left(W+50 \times 10^{-3}\right)=10\left(W+100 \times 10^{-3}\right) \\
& \Rightarrow \mathrm{W}=12.5 \times 10^{-3} \mathrm{~kg} \\
& \Rightarrow \mathrm{W}=12.5 \mathrm{~g}
\end{aligned}
\)

Hint

In steady state, the body has reached equilibrium. So, no more heat will be exchanged between the body and the surrounding.
This implies that at steady state, Rate of loss of heat \(=\) Rate at which heat is supplied
Given:
Mass, \(m=1 \mathrm{~kg}\)
Power of the heater \(=20 \mathrm{~W}\)
Room temperature \(=20^{\circ} \mathrm{C}\)
(a)At steady state, Rate of loss of heat \(=\) Rate at which heat is supplied and, rate of loss/gain of heat = Power
\(\therefore \frac{d Q}{d t}=p=20 W\)
(b) By Newton’s law of cooling, rate of cooling is directly proportional to the difference in temperature.
So, when the body is in steady state, then its rate of cooling is given as
\(
\begin{aligned}
& \frac{d Q}{d t}=K\left(T-T_0\right) 20=K(50-20) \\
& \Rightarrow K=\frac{2}{3}
\end{aligned}
\)
When the temperature of the body is \(30^{\circ} \mathrm{C}\), then its rate of cooling is given as
\(
\begin{aligned}
& \frac{d Q}{d t}=K\left(T-T_0\right) =\frac{2}{3}(30-20)=\frac{20}{3} W
\end{aligned}
\)
(c) The initial rate of cooling when the body’s temperature is \(20^{\circ} \mathrm{C}\) is given as
\(
\begin{aligned}
& \left(\frac{d Q}{d t}\right)_{20}=0 \\
& \therefore\left(\frac{d Q}{d t}\right)_{30}=\frac{20}{3} \\
& \left(\frac{d Q}{d t}\right)_{a v g}=\frac{10}{3} \\
& \mathrm{t}=5 \mathrm{~min}=300 \mathrm{~s} \\
& \text { Heat liberated }=\frac{10}{3} \times 300=1000 J \\
& \text { Net heat absorbed }=\text { Heat supplied }-\text { Heat Radiated } \\
& =6000-1000=5000 \mathrm{~J} \\
\end{aligned}
\)
d) Net heat absorbed is used for raising the temperature of the body by \(10^{\circ} \mathrm{C}\).
\(
\begin{aligned}
& \therefore m \mathrm{~S} \Delta \mathrm{T}=5000 \\
& S=\frac{5000}{m \times \Delta T}=\frac{5000}{1 \times 10} =500 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{C}^{-1} \\
\end{aligned}
\)

Hint

Heat capacity \(=\mathrm{m} \times \mathrm{s}=80 \mathrm{~J} /{ }^{\circ} \mathrm{C}\)
\(
\begin{aligned}
& \left(\frac{\mathrm{d} \theta}{\mathrm{dt}}\right)_{\text {increase }}=2^{\circ} \mathrm{C} / \mathrm{s} \\
& \left(\frac{\mathrm{d} \theta}{\mathrm{dt}}\right)_{\text {decrease }}=0.2^{\circ} \mathrm{C} / \mathrm{s}
\end{aligned}
\)
(a) Power of heater \(=\mathrm{mS}\left(\frac{\mathrm{d} \theta}{\mathrm{dt}}\right)_{\text {increa sing }}=80 \times 2=160 \mathrm{~W}\)
(b) Power radiated \(=\mathrm{ms}\left(\frac{d \theta}{d t}\right)_{\text {decrea sing }}=80 \times 0.2=16 \mathrm{~W}\)
(c) Now \(m S\left(\frac{d \theta}{d t}\right)_{\text {decreasing }}=K\left(T-T_0\right)\)
\(
\Rightarrow 16=\mathrm{K}(30-20) \quad \Rightarrow \mathrm{K}=\frac{16}{10}=1.6
\)
Now, \(\frac{d \theta}{d t}=K\left(T-T_0\right)=1.6 \times(30-25)=1.6 \times 5=8 \mathrm{~W}\)
(d)
As we know that power given by the heater is
\(
\mathrm{P}=160 \mathrm{~W}
\)
now we can use energy equations
Energy given by heater \(=\) Heat absorbed by block \(+\) heat loss \(160 t=80(30-20)+8 t\)
\(152 t=800\)
\(t=5.2 \mathrm{~s}\)

Hint

According to Newton’s law of cooling,
\(
\frac{d \theta}{d t}=-K\left(\theta-\theta_0\right)
\)
(a) Maximum heat that the body can lose, \(\Delta Q_{\max }=m s\left(\theta_1-\theta_0\right)\)
(b) If the body loses \(90 \%\) of the maximum heat, then the fall in temperature will be \(\theta\).
\(
\begin{aligned}
& \Delta Q_{\max } \times \frac{90}{100}=m s\left(\theta_1-\theta\right) \\
& \Rightarrow m s\left(\theta_1-\theta_0\right) \times \frac{9}{10}=m s\left(\theta_1-\theta\right) \\
& \Rightarrow \theta=\theta_1-\left(\theta_1-\theta_0\right) \times \frac{9}{10} \\
& \Rightarrow \theta=\frac{\theta_1-9 \theta_0}{10} \dots(i)
\end{aligned}
\)
From Newton’s law of cooling,
\(
\frac{d \theta}{d t}=-K\left(\theta_1-\theta\right)
\)
Integrating this equation within the proper limit, we get
At time \(t=0\),
\(
\theta=\theta_1
\)
At time \(t\)
\(
\begin{aligned}
& \theta=\theta \\
& \int_{\theta 1}^\theta \frac{d \theta}{\theta_1-\theta}=-K \int_0^t d t \\
& \Rightarrow \operatorname{ln} \frac{\theta_1-\theta}{\theta_1-\theta_0}=-k t \\
& \Rightarrow \theta_1-\theta=\theta_1-\theta_0 e^{-k t} \dots(ii)
\end{aligned}
\)
From (i) and (ii),
\(
\begin{aligned}
& \frac{\theta_1-9 \theta_0}{10}-\theta_0=\left(\theta_1-\theta_0\right) e^{-k} t \\
& \Rightarrow t=\frac{\operatorname{In}(10)}{k}
\end{aligned}
\)