We have seen that heat is energy transfer from one system to another or from one part of a system to another part, arising due to temperature differences. What are the different ways by which this energy transfer takes place? There are three distinct modes of heat transfer:

• Conduction
• Convection
• Radiation

This is shown in Fig 11.13 below.

Conduction

Conduction is the mechanism of transfer of heat between two adjacent parts of a body because of their temperature difference. Suppose, one end of a metallic rod is put in a flame, the other end of the rod will soon be so hot that you cannot hold it by your bare hands. Here, heat transfer takes place by conduction from the hot end of the rod through its different parts to the other end. Gases are poor thermal conductors, while liquids have conductivities intermediate between solids and gases.

Heat conduction may be described quantitatively as the time rate of heat flow in a material for a given temperature difference. Consider a metallic bar of length \(L\) and uniform cross-section \(A\) with its two ends maintained at different temperatures. This can be done, for example, by putting the ends in thermal contact with large reservoirs at temperatures, say, \(T_{\mathrm{C}}\) and \(T_{\mathrm{D}}\), respectively (Flg. 11.14). Let us assume the ideal condition that the sides of the bar are fully insulated so that no heat is exchanged between the sides and the surroundings.

After sometime, a steady state is reached; the temperature of the bar decreases uniformly with distance from \(T_{\mathrm{C}}\) to \(T_{\mathrm{D}} ;\left(T_{\mathrm{C}}>T_{\mathrm{D}}\right)\). The reservoir at C supplies heat at a constant rate, which transfers through the bar and is given out at the same rate to the reservoir at D. It is found experimentally that in this steady state, the rate of flow of heat (or heat current) \(H\) is proportional to the temperature difference \(\left(T_{\mathrm{C}}-T_{\mathrm{D}}\right)\) and the area of cross-section \(A\) and is inversely proportional to the length \(L\) :
\(
H=\frac{\Delta Q}{\Delta t}=K A \frac{T_C-T_D}{L} \dots(11.14)
\)
The constant of proportionality \(K\) is called the thermal conductivity of the material. The greater the value of \(K\) for a material, the more rapidly will it conduct heat. \(\Delta Q\) is the amount of heat crosses through any cross-section in time \(\Delta t, \Delta Q / \Delta t\) is called the heat current.

The SI unit of \(K\) is \(\mathrm{J}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}\) or \(\mathrm{W} \mathrm{m}^{-1} \mathrm{~K}^{-1}\). 

In some books, the above equation is written as \(\frac{\Delta Q}{\Delta t}=K \frac{A\left(T_1-T_2\right)}{x}\). They are basically the same except variables are taken differently like \(L\) = \(x\). 

The thermal conductivities of various substances are listed in Table 11.6. These values vary slightly with temperature but can be considered to be constant over a normal temperature range.

Example 1: What is the temperature of the steel-copper junction in the steady state of the system shown in Fig. 11.15? Length of the steel rod \(=15.0 \mathrm{~cm}\), length of the copper rod \(=10.0 \mathrm{~cm}\), temperature of the furnace \(=300^{\circ} \mathrm{C}\), temperature of the other end \(=0^{\circ} \mathrm{C}\). The area of cross-section of the steel rod is twice that of the copper rod. (Thermal conductivity of steel \(=50.2 \mathrm{~J} \mathrm{~s}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}\); and of copper \(\left.=385 \mathrm{~J} \mathrm{~s}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}\right)\)

Solution:

The insulating material around the rods reduces heat loss from the sides of the rods. Therefore, heat flows only along the length of the rods. Consider any cross-section of the rod. In the steady state, heat flowing into the element must equal the heat flowing out of it; otherwise, there would be a net gain or loss of heat by the element and its temperature would not be steady. Thus in the steady state, the rate of heat flowing across a cross-section of the rod is the same at every point along the length of the combined steel-copper rod.

Let \(T\) be the temperature of the steel-copper junction in the steady state. Then,
\(
\frac{K_1 A_1(300-T)}{L_1}=\frac{K_2 A_2(T-0)}{L_2}
\)
where 1 and 2 refer to the steel and copper rod respectively. For \(A_1=2 A_2, L_1=15.0 \mathrm{~cm}\), \(L_2=10.0 \mathrm{~cm}, K_1=50.2 \mathrm{~J} \mathrm{~s}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}, K_2=385 \mathrm{~J}\) \(\mathrm{s}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}\), we have
\(
\frac{50.2 \times 2(300-T)}{15}=\frac{385 T}{10}
\)
which gives \(T=44.4^{\circ} \mathrm{C}\)

Example 2: An iron bar \(\left(L_1=0.1 \mathrm{~m}, A_1\right.\) \(\left.=0.02 \mathrm{~m}^2, K_1=79 \mathrm{~W} \mathrm{~m}{ }^{-1} \mathrm{~K}^{-1}\right)\) and \(\mathrm{a}\) brass bar \(\left(L_2=0.1 \mathrm{~m}, A_2=0.02 \mathrm{~m}^2\right.\), \(K_2=109 \mathrm{~W} \mathrm{~m}^{-1} \mathrm{~K}^{-1}\) ) are soldered end to end as shown in Fig. 11.16. The free ends of the iron bar and brass bar are maintained at \(373 \mathrm{~K}\) and \(273 \mathrm{~K}\) respectively. Obtain expressions for and hence compute (i) the temperature of the junction of the two bars, (ii) the equivalent thermal conductivity of the compound bar, and (i11) the heat current through the compound bar.

Solution:

Given, \(L_1=L_2=L=0.1 \mathrm{~m}, A_1=A_2=A=0.02 \mathrm{~m}^2\)
\(K_1=79 \mathrm{~W}^{-1} \mathrm{~K}^{-1}, K_2=109 \mathrm{~W} \mathrm{~m}^{-1} \mathrm{~K}^{-1}\),
\(T_1=373 \mathrm{~K}\), and \(T_2=273 \mathrm{~K}\).
Under steady state condition, the heat current \(\left(H_1\right)\) through iron bar is equal to the heat current \(\left(\mathrm{H}_2\right)\) through brass bar.
So, \(H=H_1=H_2\)
\(
=\frac{K_1 A_1\left(T_1-T_0\right)}{L_1}=\frac{K_2 A_2\left(T_0-T_2\right)}{L_2}
\)
For \(A_1=A_2=A\) and \(L_1=L_2=L\), this equation leads to
\(
K_1\left(T_1-T_0\right)=K_2\left(T_0-T_2\right)
\)
Thus, the junction temperature \(T_0\) of the two bars is
\(
T_0=\frac{\left(K_1 T_1+K_2 T_2\right)}{\left(K_1+K_2\right)}
\)
Using this equation, the heat current \(H\) through either bar is
\(
\begin{aligned}
H & =\frac{K_1 A\left(T_1-T_0\right)}{L}=\frac{K_2 A\left(T_0-T_2\right)}{L} \\
& =\left(\frac{K_1 K_2}{K_1+K_2}\right) \frac{A\left(T_1-T_0\right)}{L}=\frac{A\left(T_1-T_2\right)}{L\left(\frac{1}{K_1}+\frac{1}{K_2}\right)}
\end{aligned}
\)
Using these equations, the heat current \(H^{\prime}\) through the compound bar of length \(L_1+L_2=2 L\) and the equivalent thermal conductivity \(K^{\prime}\), of the compound bar are given by
\(
\begin{aligned}
& H^{\prime}=\frac{K^{\prime} A\left(T_1-T_2\right)}{2 L}=H \\
& K^{\prime}=\frac{2 K_1 K_2}{K_1+K_2}
\end{aligned}
\)
(i)
\(
\begin{aligned}
& T_0=\frac{\left(K_1 T_1+K_2 T_2\right)}{\left(K_1+K_2\right)} \\
& =\frac{\left(79 \mathrm{~W} \mathrm{~m}^{-1} \mathrm{~K}^{-1}\right)(373 \mathrm{~K})+\left(109 \mathrm{~W} \mathrm{~m}^{-1} \mathrm{~K}^{-1}\right)(273 \mathrm{~K})}{79 \mathrm{~W} \mathrm{~m}^{-1} \mathrm{~K}^{-1}+109 \mathrm{~W}{ }^{-1} \mathrm{~K}^{-1}} \\
& =315 \mathrm{~K}
\end{aligned}
\)
(ii)
\(
\begin{aligned}
K^{\prime} & =\frac{2 K_1 K_2}{K_1+K_2} \\
& =\frac{2 \times\left(79 \mathrm{~W} \mathrm{~m}^{-1} \mathrm{~K}^{-1}\right) \times\left(109 \mathrm{~W} \mathrm{~m}^{-1} \mathrm{~K}^{-1}\right)}{79 \mathrm{~W} \mathrm{~m}^{-1} \mathrm{~K}^{-1}+109 \mathrm{~W} \mathrm{~m}^{-1} \mathrm{~K}^{-1}} \\
& =91.6 \mathrm{~W} \mathrm{~m}^{-1} \mathrm{~K}^{-1}
\end{aligned}
\)
\(
\begin{aligned}
& \text { (iii) } H^{\prime}=H=\frac{K^{\prime} A\left(T_1-T_2\right)}{2 L} \\
& =\frac{\left(91.6 \mathrm{~W} \mathrm{~m}^{-1} \mathrm{~K}^{-1}\right) \times\left(0.02 \mathrm{~m}^2\right) \times(373 \mathrm{~K}-273 \mathrm{~K})}{2 \times(0.1 \mathrm{~m})} \\
& =916.1 \mathrm{~W} \\
&
\end{aligned}
\)

Example 3: One face of a copper cube of edge \(10 \mathrm{~cm}\) is maintained at \(100^{\circ} \mathrm{C}\) and the opposite face is maintained at \(0^{\circ} \mathrm{C}\). All other surfaces are covered with an insulating material. Find the amount of heat flowing per second through the cube. Thermal conductivity of copper is \(385 \mathrm{Wm}^{-1_0} \mathrm{C}^{-1}\).

Solution:

The heat flows from the hotter face towards the colder face. The area of cross-section perpendicular to the heat flow is
\(
A=(10 \mathrm{~cm})^2 .
\)
The amount of heat flowing per second is
\(
\begin{aligned}
\frac{\Delta Q}{\Delta t} & =K A \frac{T_1-T_2}{L} \\
& =\left(385 \mathrm{Wm}^{-10} \mathrm{C}^{-1}\right) \times(0.1 \mathrm{~m})^2 \times \frac{\left(100^{\circ} \mathrm{C}-0^{\circ} \mathrm{C}\right)}{0.1 \mathrm{~m}} \\
& =3850 \mathrm{~W} .
\end{aligned}
\)

Thermal Resistance

In the equation \(\frac{\Delta Q}{\Delta t}=K \frac{A\left(T_1-T_2\right)}{x}\), the quantity \(\frac{x}{K A}\) is called the thermal resistance \(R\). Writing the heat current \(\Delta Q / \Delta t\) as \(i\), we have
\(
i=\frac{T_1-T_2}{R} .
\)
This is mathematically equivalent to Ohm’s law to be introduced in a later chapter.

Example 4: Find the thermal resistance of an aluminium rod of length \(20 \mathrm{~cm}\) and area of cross-section \(1 \mathrm{~cm}^2\). The heat current is along the length of the rod. Thermal conductivity of aluminium \(=200 \mathrm{Wm}^{-1} \mathrm{~K}^{-1}\).

Solution:

The thermal resistance is
\(
R=\frac{x}{K A}=\frac{20 \times 10^{-2} \mathrm{~m}}{\left(200 \mathrm{Wm}^{-1} \mathrm{~K}^{-1}\right)\left(1 \times 10^{-4} \mathrm{~m}^2\right)}=10 \mathrm{KW}^{-1}
\)

Series Connection Of Rods

Consider two rods of thermal resistances \(R_1\) and \(R_2\) joined one after the other as shown in figure above. The free ends are kept at temperatures \(T_1\) and \(T_2\) with \(T_1>T_2\). In steady state, any heat that goes through the first rod also goes through the second rod. Thus, the same heat current passes through the two rods. Such a connection of rods is called a series connection. Suppose, the temperature of the junction is \(T\). Using the heat current equation we can write the heat current through the first rod is
\(
i=\frac{\Delta Q}{\Delta t}=\frac{T_1-T}{R_1}
\)
or, \(\quad T_1-T=R_1 i \dots(i)\)
and that through the second rod is
\(
i=\frac{\Delta Q}{\Delta t}=\frac{T-T_2}{R_2}
\)
or, \(\quad T-T_2=R_2 i \dots(ii)\).
Adding (i) and (ii),
\(
\begin{aligned}
T_1-T_2 & =\left(R_1+R_2\right) i \\
i & =\frac{T_1-T_2}{R_1+R_2} .
\end{aligned}
\)
Thus, the two rods together is equivalent to a single rod of thermal resistance \(R_1+R_2\).

If more than two rods are joined in series, the equivalent thermal resistance is given by
\(
R=R_1+R_2+R_5+\ldots
\)

Parallel Connection Of Rods

Now, suppose the two rods are joined at their ends as shown in figure above. The left ends of both the rods are kept at temperature \(T_1\) and the right ends are kept at temperature \(T_2\). So the same temperature difference is maintained between the ends of each rod. Such a connection of rods is called a parallel connection. The heat current going through the first rod is
\(
i_1=\frac{\Delta Q_1}{\Delta t}=\frac{T_1-T_2}{R_1}
\)
and that through the second rod is
\(
i_2=\frac{\Delta Q_2}{\Delta t}=\frac{T_1-T_2}{R_2} \text {. }
\)
The total heat current going through the left end is
\(
\begin{array}{rlrl}
i =i_1+i_2 \\
 =\left(T_1-T_2\right)\left(\frac{1}{R_1}+\frac{1}{R_2}\right) \\
\text { or, } i  =\frac{T_1-T_2}{R} \\
\text { where } \frac{1}{R} =\frac{1}{R_1}+\frac{1}{R_2}  \dots(i)
\end{array}
\)

Thus, the system of the two rods is equivalent to a single rod of thermal resistance \(R\) given by (i).
If more than two rods are joined in parallel, the equivalent thermal resistance \(R\) is given by
\(
\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\ldots
\)

What is Convection?

Convection is the process of heat transfer by the bulk movement of molecules within fluids such as gases and liquids. The initial heat transfer between the object and the fluid takes place through conduction, but the bulk heat transfer happens due to the motion of the fluid.

• Convection is the process of heat transfer in fluids by the actual motion of matter.
• It happens in liquids and gases.
• It may be natural or forced.
• It involves a bulk transfer of portions of the fluid.

How is Heat Transferred through Convection?

When a fluid is heated from below, thermal expansion takes place. The lower layers of the fluid, which are hotter, become less dense. We know that colder fluid is denser. Due to buoyancy, the less dense, hotter part of the fluid rises up. And the colder, denser fluid replaces it. This process is repeated when this part also gets heated and rises up to be replaced by the colder upper layer. This is how the heat is transferred through convection.

Types of Convection

There are two types of convection, and they are:

• Natural convection
• Forced convection

Natural convection: When convection takes place due to buoyant force as there is a difference in densities caused by the difference in temperatures it is known as natural convection. Examples of natural convection are oceanic winds.

Forced convection: When external sources such as fans and pumps are used for creating induced convection, it is known as forced convection. Examples of forced convection are using water heaters or geysers for instant heating of water and using a fan on a hot summer day.

Natural convection is responsible for many familiar phenomena. During the day, the ground heats up more quickly than large bodies of water do. This occurs both because water has a greater specific heat capacity and because mixing currents disperse the absorbed heat throughout the great volume of water. The air in contact with the warm ground is heated by conduction. It expands, becoming less dense than the surrounding cooler air. As a result, the warm air rises (air currents) and the other air moves (winds) to fill the space-creating a sea breeze near a large body of water. Cooler air descends, and a thermal convection cycle is set up, which transfers heat away from the land.  At night, the ground loses its heat more quickly, and the water surface is warmer than the land. As a result, the cycle is reversed (Fig. 11.17).

Radiation

Radiation is defined as a process involving the transfer of heat from one place to another place without heating the intervening medium. it is called radiation and the energy so transferred by electromagnetic waves is called radiant energy. These waves are formed due to the superposition of electric and magnetic fields perpendicular to each other and carry energy.

In an electromagnetic wave, electric and magnetic fields oscillate in space and time. Like any wave, electromagnetic waves can have different wavelengths and can travel in vacuum with the same speed, namely the speed of light i.e., \(3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\). You will learn these matters in more detail later, but you now know why heat transfer by radiation does not need any medium and why it is so fast. This is how heat is transferred to the earth from the Sun through empty space. All bodies emit radiant energy, whether they are solid, liquid or gas. The electromagnetic radiation emitted by a body by virtue of its temperature, like radiation by a red-hot iron or light from a filament lamp is called thermal radiation.

Important Types Of Radiation

In Physics, we study different forms of Radiation. The commons ones include;

• Electromagnetic radiation: These include radiations such as microwaves, infrared, ultraviolet, radio waves, x-rays, and gamma radiation ( \(\mathrm{y}\) ), and visible light.
• Particle radiation: These include alpha radiation ( \(\alpha\) ), beta radiation ( \(\beta\) ), and neutron radiation.
• Acoustic radiation: Some of the popular ones are sound, ultrasound, and seismic waves.
• Gravitational radiation: This is a type of radiation that often takes the form of gravitational waves or ripples in the curvature of spacetime.
  Additionally, radiation is mainly described or divided as ionizing or non-ionizing radiation. This is done depending on the energy of the radiated particles.

Properties of Radiation

• All objects emit radiation simply because their temperature is above absolute zero, and all objects absorb some of the radiation that falls on them from other objects.
• Maxwell on the basis of his electromagnetic theory proved that all radiations are electromagnetic waves and their sources are vibrations of charged particles in atoms and molecules.
• More radiation are emitted at a higher temperature of a body and lesser at a lower temperature.
• The wavelength corresponding to the maximum emission of radiations shifts from longer wavelength to a shorter wavelength as the temperature increases. Due to this, the colour of a body appears to be changing. Radiations from a body at NTP has predominantly infrared waves.
• Thermal radiations travel with the speed of light and in a straight line.
• Radiations are electromagnetic waves and can also travel through a vacuum.
• Similar to light, thermal radiations can be reflected, refracted, diffracted, and polarized.
• Radiation from a point source obeys the inverse square law (intensity a \(1 / \mathrm{r}^2\) ).

Blackbody Radiation

A perfectly black body is one which absorbs all the heat radiations of whatever wavelength incident on it. It neither reflects nor transmits any of the incident radiation and therefore appears black, whatever the colour of the incident radiation.

In actual practice, no natural object possesses strictly the properties of a perfectly black body. But the lamp-black and platinum black are a good approximation of black body. They absorb about 99 % of the incident radiation. 

The important thing about thermal radiation at any temperature is that it is not of one (or a few) wavelength(s) but has a continuous spectrum from the small to the long wavelengths. The energy content of radiation, however, varies for different wavelengths. Figure 11.18 gives the experimental curves for radiation energy per unit area per unit wavelength emitted by a blackbody versus wavelength for different temperatures.

Notice that the wavelength \(\lambda_m\) for which energy is the maximum decreases with increasing temperature.

Wien’s Displacement Law

The relation between \(\lambda_m\) and \(T\) is given by what is known as Wien’s Displacement Law:
\(
\lambda_m T=\text { constant } \dots(11.15)
\)
The value of the constant (Wien’s constant) is \(2.9 \times 10^{-3} \mathrm{~m} \mathrm{~K}\). This law explains why the colour of a piece of iron heated in a hot flame first becomes dull red, then reddish yellow, and finally white hot. Wien’s law is useful for estimating the surface temperatures of celestial bodies like the moon, Sun, and other stars. Light from the moon is found to have a maximum intensity near the wavelength \(14 \mu \mathrm{m}\). By Wien’s law, the surface of the moon is estimated to have a temperature of \(200 \mathrm{~K}\). Solar radiation has a maximum at \(\lambda_m=4753 \AA\). This corresponds to \(T=6060 \mathrm{~K}\). Remember, this is the temperature of the surface of the sun, not it’s interior.

Example 5: The light from the sun is found to have a maximum intensity near the wavelength of \(470 \mathrm{~nm}\). Assuming that the surface of the sun emits as a blackbody, calculate the temperature of the surface of the sun.

Solution:

For a blackbody, \(\lambda_m T=0.288 \mathrm{cmK}\).
\(
\text { Thus, } T=\frac{0.288 \mathrm{cmK}}{470 \mathrm{~nm}}=6130 \mathrm{~K} \text {. }
\)

Stefan-Boltzmann law

Energy can be transferred by radiation over large distances, without a medium (1.e., in vacuum). The total electromagnetic energy radiated by a body at absolute temperature \(T\) is proportional to its size, its ability to radiate (called emissivity) and most importantly to its temperature. For a body, which is a perfect radiator, the energy emitted per unit time \((H)\) is given by
\(
H=A \sigma T^4 \dots(11.16)
\)
where \(A\) is the area and \(T\) is the absolute temperature of the body. This relation obtained experimentally by Stefan and later proved theoretically by Boltzmann is known as the Stefan-Boltzmann law and the constant \(\sigma\) is called the Stefan-Boltzmann constant. Its value in SI units is \(5.67 \times 10^{-8} \mathrm{~W} \mathrm{~m}^{-2} \mathrm{~K}^{-4}\). Most bodies emit only a fraction of the rate given by Eq. 11.16. A substance like lamp black comes close to the limit. One, therefore, defines a dimensionless fraction \(e\) called emissivity and writes,
\(
H=A e \sigma T^4 \dots(11.17)
\)
Here, \(e=1\) for a perfect radiator. For a tungsten lamp, for example, \(e\) is about \(0.4\). Thus, a tungsten lamp at a temperature of \(3000 \mathrm{~K}\) and a surface area of \(0.3 \mathrm{~cm}^2\) radiates at the rate \(H=0.3 \times\) \(10^{-4} \times 0.4 \times 5.67 \times 10^{-8} \times(3000)^4=60 \mathrm{~W}\).

A body at temperature \(T\), with surroundings at temperatures \(T_s\), emits, as well as, receives energy. For a perfect radiator, the net rate of loss of radiant energy is
\(
H=\sigma A\left(T^4-T_s^4\right)
\)
For a body with emissivity \(e\), the relation modifies to
\(
H=e \sigma A\left(T^4-T_s^4\right) \dots(11.18)
\)
As an example, let us estimate the heat radiated by our bodies. Suppose the surface area of a person’s body is about \(1.9 \mathrm{~m}^2\) and the room temperature is \(22^{\circ} \mathrm{C}\). The internal body temperature, as we know, is about \(37^{\circ} \mathrm{C}\). The skin temperature may be \(28^{\circ} \mathrm{C}\) (say). The emissivity of the skin is about \(0.97\) for the relevant region of electromagnetic radiation. The rate of heat loss is:
\(
\begin{aligned}
H & =5.67 \times 10^{-8} \times 1.9 \times 0.97 \times\left\{(301)^4-(295)^4\right\} \\
& =66.4 \mathrm{~W}
\end{aligned}
\)
which is more than half the rate of energy production by the body at rest \((120 \mathrm{~W})\). To prevent this heat loss effectively (better than ordinary clothing), modern arctic clothing has an additional thin shiny metallic layer next to the skin, which reflects the body’s radiation.

Example 6: A blackbody of the surface area \(10 \mathrm{~cm}^2\) is heated to \(127^{\circ} \mathrm{C}\) and is suspended in a room at a temperature \(27^{\circ} \mathrm{C}\). Calculate the initial rate of loss of heat from the body to the room.

Solution:

For a blackbody at temperature \(T\), the rate of emission is \(u=\sigma A T^4\). When it is kept in a room at temperature \(T_0\), the rate of absorption is \(u_0=\sigma A T_0^4\).
The net rate of loss of heat is \(u-u_0=\sigma A\left(T^4-T_0^4\right)\).
Here \(A=10 \times 10^{-4} \mathrm{~m}^2, T=400 \mathrm{~K}\) and \(T_0=300 \mathrm{~K}\). Thus,
\(
\begin{aligned}
& u-u_0 \\
& =\left(5.67 \times 10^{-8} \mathrm{Wm}^{-2} \mathrm{~K}^{-4}\right)\left(10 \times 10^{-4} \mathrm{~m}^2\right)\left(400^4-300^4\right) \mathrm{K}^4 \\
& =0.99 \mathrm{~W} .
\end{aligned}
\)

Kirchhoff’s Law

LawThe ratio of emissive power to absorptive power is the same for all bodies at a given temperature and is equal to the emissive power of a blackbody at that temperature. Thus,

\(
\frac{E(\text { body })}{a(\text { body })}=E(\text { blackbody })
\)

Where the absorptive power \( a=\frac{\text { energy absorbed }}{\text { energy incident }}\) and emissive power of the body as

\(E=\frac{\Delta U}{(\Delta A)(\Delta \omega)(\Delta t)}\)

Greenhouse Effect

The earth’s surface is a source of thermal radiation as it absorbs energy received from the Sun. The wavelength of this radiation lies in the long wavelength (infrared) region. But a large portion of this radiation is absorbed by greenhouse gases, namely, carbon dioxide \(\left(\mathrm{CO}_2\right)\); methane \(\left(\mathrm{CH}_4\right)\); nitrous oxide \(\left(\mathrm{N}_2 \mathrm{O}\right)\); chlorofluorocarbon \(\left(\mathrm{CF}_{\mathrm{x}} \mathrm{Cl}_{\mathrm{x}}\right)\); and tropospheric ozone \(\left(\mathrm{O}_3\right)\). This heats up the atmosphere which, in turn, gives more energy to the earth, resulting in warmer surface. This increases the intensity of radiation from the surface. The cycle of processes described above is repeated until no radiation is available for absorption. The net result is heating up of the earth’s surface and atmosphere. This is known as Greenhouse Effect. Without the Greenhouse Effect, the temperature of the earth would have been \(-18^{\circ} \mathrm{C}\).

Concentration of greenhouse gases has enhanced due to human activities, making the earth warmer. According to an estimate, the average temperature of the earth has increased by \(0.3\) to \(0.6^{\circ} \mathrm{C}\), since the beginning of this century because of this enhancement. By the middle of the next century, the earth’s global temperature may be 1 to \(3^{\circ} \mathrm{C}\) higher than today. This global warming may cause problems for human life, plants, and animals. Because of global warming, ice caps are melting faster, sea level is rising, and weather pattern is changing. Many coastal cities are at the risk of getting submerged.

Short Questions

Q1. The heat current is written as \(\frac{\Delta Q}{\Delta t}\). Why don’t we write \(\frac{d Q}{d t}?\)

Answer: The amount of heat crossing through any cross-section of a slab in time \(\Delta t\) is called heat current. It is written as \(\frac{\Delta Q}{\Delta t}\) and not \(\frac{\Delta Q}{\Delta t}\) as complete derivative. This is because the amount of heat crossing through any cross-section is a function of many variables like temperature difference, area of cross-section, etc. So, we cannot write it as a complete derivative with respect to time.

Q2. Does a body at \(20^{\circ} \mathrm{C}\) radiate in a room, where the room temperature is \(30^{\circ} \mathrm{C}\)? If yes, why does its temperature not fall further?

Answer: Yes, the body will radiate. However, its temperature will not fall down with time because as the temperature of the surroundings is greater than the temperature of the body so, its rate of absorption will be greater than its rate of emission.

Q3. Why does blowing over a spoonful of hot tea cools it ? Does evaporation play a role? Does radiation play a role?

Answer: Here, major role is played by convection. When we blow air over a spoonful of hot tea, the air coming from our mouth has less temperature than the air above the tea. Since hot air has less density, it rises up and cool air goes down. In this way, the tea cools down. We know that any hot body radiates. So, the spoonful of tea will also radiate and as the temperature of the surrounding is less than the tea, the tea will cool down with time. Evaporation is also involved in this. On blowing over the hot tea, the rate of evaporation increases and the cools down.

Q4. On a hot summer day we want to cool our room by opening the refrigerator door and closing all the windows and doors. Will the process work?

Answer: No. When the door of the refrigerator is left open in a closed room, the heat given out by the refrigerator to the room will be more than that taken from the room. Therefore, instead of decreasing, the temperature of the room will increase at a slower rate.

Q5. On a cold winter night, you are asked to sit on a chair. Would you like to choose a metal chair or a wooden chair? Both are kept in the same lawn and are at the same temperature.

Answer: We will prefer to seat on a wooden chair because as the conductivity of wood is poorer than that of metal, heat flow from our body to the chair will be less in the case of a wooden chair.

Q6. Two identical metal balls one at \(T_1=300 \mathrm{~K}\) and the other at \(T_2=600 \mathrm{~K}\) are kept at a distance of \(1 \mathrm{~m}\) in vacuum. Will the temperatures equalise by radiation? Will the rate of heat gained by the colder sphere be proportional to \(T_2^4-T_1^4\) as may be expected from the Stefan’s law?

Answer: Yes, the temperature of the balls can be equalised by radiation. This is because both spheres will emit radiations in all directions at different rates.

The ball kept at the temperature of \(300 \mathrm{~K}\) will gain some thermal energy by the radiation emitted by the ball kept at the temperature of \(600 \mathrm{~K}\). Also, it losses energy by radiation. Similarly, the ball kept at the temperature of \(600 \mathrm{~K}\) will gain some thermal energy by the radiation emitted by the ball kept at the temperature of \(600 \mathrm{~K}\). Also, it losses energy by radiation. A time comes when the temperature of both bodies becomes equal. Yes, the rate of heat gained by the colder sphere is proportional to \(t_2^4-t_1^4\).

Q7. An ordinary electric fan does not cool the air, still, it gives comfort in summer. Explain?

Answer: An ordinary electric fan does not cool the air, but still it gives comfort in summer because it circulates the air present in the room. Due to this, evaporation takes place and we feel cooler.

Q8. The temperature of the atmosphere at a high altitude is around \(500^{\circ} \mathrm{C}\). Yet an animal there would freeze to death and not boil. Explain.

Answer: The temperature of the atmosphere at a high altitude is around \(500^{\circ} \mathrm{C}\), but density of air molecule is extremely low at this height. So, very less molecules of air collide with the body of the animal and transfer very less amount of heat. That is why the animal present there would freeze to death instead of boiling.

Q9. Standing in the sun is more pleasant on a cold winter day than standing in shade. Is the temperature of air in the sun considerably higher than that of the air in shade?

Answer: The heat coming from the sun to us is through radiation. On colder winter days, when we stand in shade, we do not get the heat of the sun from the radiation. Though we feel cool in the shade, the temperature of the air in shady as well as non-shady regions is the same.

Q10. Cloudy nights are warmer than the nights with clean sky. Explain?

Answer: During night, the earth’s surface radiates infrared radiation of larger wavelength. Gas molecules in the air absorb some of this energy and radiate energy of their own in all directions. Also, water molecules, like the vapour that makes the clouds, absorb more frequencies of infrared energy than clear air does.

Both these factors contribute to the fact that clouds radiate more heat in all directions (including the earth) than clear air does. In turn, this makes the overall temperature on the earth warmer when there is a cloud cover. The heat energy radiated by the earth is reflected back to earth. Due to this, cloudy nights are warmer than the nights with clean sky.

Q11. Why is a white dress more comfortable than a dark dress in summer?

Answer: A white colour dress reflects almost all the radiations falling on it. So, it does not absorb any heat from the sunlight and we feel more comfortable in it. On the other hand, a dark colour dress absorbs maximum radiation falling on it. So, we feel hot in a dark-coloured dress during summers.