Topics

Radiation is the process in which energy is transferred by means of electromagnetic waves.
All bodies continuously radiate energy in the form of electromagnetic waves. It does not require a material medium. Electromagnetic waves from the sun, for example, travel through the void of space during their journey to earth. Even an ice cube radiates energy, although so little of it is in the form of visible light that an ice cube cannot be seen in the dark. The surface of an object plays a significant role in determining how much radiant energy the object will absorb or emit.

Figure : The temperature of the block coated with lampblack rises faster than the temperature of the block coated absorbs radiant energy from the sun at the greater rate

The two blocks in sunlight in figure, for example, are identical, except that one has a rough surface coated with lamblack (a fine black soot), while the other has a highly polished silver surface. As the thermometers indicate, the temperature of the black block rises at a much faster rate than that of the silver block. This is because lampblack absorbs about \(97\%\) of the incident radiant energy, while the silvery surface absorbs only about \(10\%\). The remaining part of the incident energy is reflected in each case. We observe the lampblack as black in color because it reflects so little of the light falling on it, while the silvery surface looks like a mirror because it reflects so little of the light falling on it, while the silvery surface looks like a mirror because it reflects so much light. Since the color black is associated with nearly complete absorption of visible light, the term perfect blackbody or, simply, blackbody is used when referring to an object that absorbs all the electromagnetic waves falling on it.

Black body:

The experiments described before lead us to the idea of a perfectly black body, one which absorbs all the radiation that falls upon it, and reflects and transmits none. The experiments also lead us to suppose that such a body would be the best possible radiator.

Prevost's theory of exchange:

Any body having temperature greater then zero kelvin, must emit or aborb radiation.

A is placed in an evacuated enclosure B, at lower temperature than A, then A cools until it reaches the temperature of B . If a body C , cooler than B , is put in B , then C warms up to the temperature of B . We conclude that radiation from \(B\) falls on \(C\), and therefore also on \(A\), even through \(A\) is at a higher temperature. Thus A and C each come to equilibrium at the temperature of B when each is absorbing and emitting radiation at equal rates.

If Q is the total incident energy on a body, \(Q_{1}\) is the part absorbed, \(Q_{2}\) is the part reflected and \(Q_{3}\) is the part transmitted then
\[Q = Q_{1} + Q_{2} + Q_{3} \]Absorption coefficient or absorptive power
Reflection coefficient
\[a = Q_{1}/Q \]Transmission coefficient
\[r = Q_{2}/Q\]

Thus \(a + r + t = 1\)
\[t = Q_{3}/Q\]

If, for a body, \(r = t = 0\) and \(a = 1\), i.e. it absorbs all the energy falling on it, such bodies are known as black bodies.

Emissive Power:

Emissive power of a surface is the quantity of heat energy emitted per second, per unit area of surface through unit solid angle. It depends on the nature and the temperature of the surface.

Emissivity:

Emissivity of a surface is the ratio of the emissive power of that surface to the emissive power of a black body at the same temperature.

Kirchhoff's Law:

At a given temperature, the ratio of emissive power to absorptive power of any body is equal to the emissive power of a black body at that temperature. Thus,
\[\frac{E_{1}}{a_{1}} = \frac{E_{2}}{a_{2}} = E_{\text{Black body}\text{~}} \]From Kirchhoff's law, it can be deduced that good absorbers are also good emitters

Stefan's radiation law

An idealized body that absorbs all the radiation incident upon it is called a blackbody. A blackbody absorbs not only all visible light, but infrared, ultraviolet, and all other wavelengths of electromagnetic radiation. It turns out that a good absorber is also a good emitter of radiation. A blackbody emits more radiant ower per unit surface area than any real object at the same temperature. The rate at which a blackbody emits rdiation per unit surface area is proportional to the fourth power of the absolute temperature.

\[P = \frac{dQ}{dt} = \sigma{AT}^{4}\ (\text{~}\text{for a black body}\text{~})\]

In equation, A is the surface area and T is the surface temperature of the blackbody in kelvins. Since Stefan's law involves the absolute temperature and not a temperature difference, \(\ ^{\circ}C\) cannot be substituted. The universal constant \(\sigma\) (Greek letter sigma) is called Stefan's constant :

\[\sigma = 5.670 \times 10^{- 8}\text{ }W/\left( m^{2} \cdot {\text{ }K}^{4} \right)\]

The fourth-power temperature dependence implies that the power emitted is extremely sensitive to temperature changes. If the absolute temperature of a body doubles, the energy emitted increases by a factor of \(2^{4} = 16\).

Since real bodies are not perfect absorbers and therefore emit less than a blackbody, we define the emissivity (e) as the ratio of the emitted power of the body to that of a blackbody at the same temperature. Then Stefan's law becomes.

\[P = e\sigma{AT}^{4}\ \text{~}\text{(for a no}\text{n-black body)}\text{~}\]

The emissivity ranges from 0 to 1 .
\(e = 1\) for a perfect radiator and absorber (a blackbody).
\(e = 0\) for a perfect reflector.

Hot object placed in isothermal enclosure:

Consider a body at a temperature of \(T_{0}\) and \(T_{e}\) is the temperature of the room or enclosure containing the body. If A is the surface area of the body and emissivity (e).
Since the body is in temperature equilibrium, the energy per second it radiates must equal the energy per second it absorbs. then, from Stefan's law,
energy per second emitted \(\left( P_{\text{emit}\text{~}} \right) = e\sigma{AT}_{0}\ ^{4}\)
energy per second absorbed \(\left( P_{\text{absorbed}\text{~}} \right) = e\sigma{AT}_{e}\ ^{4}\)
\[P_{\text{emit}\text{~}} = P_{\text{aborbed}\text{~}} \Rightarrow T_{e} = T_{\text{o}\text{~}} \]Now suppose the body X is heated electrically by a heater of power W watts and finally reaches a constant temperature T. In this case, from Prevost's theory, energy per second from heater, \(W =\) net energy per second radiated by X The net energy per second radiated by \(X = e\sigma{AT}^{4} - e\sigma{AT}_{0}\ ^{4}\). So

\[W = e\sigma{AT}^{4} - e\sigma{AT}_{0}^{4} = e\sigma\text{ }A\left( {\text{ }T}^{4} - T_{0}^{4} \right)\]

For small temperature differences, the rate of cooling, due to conduction, convection, and radiation combined, is proportional to the difference in temperature. It is a valid approximation in the transfer of heat from a radiator to a room, the loss of heat through the wall of a room, or the cooling of a cup of tea on the table.

Figure : Verification of Newton's Law of cooling

Suppose, a body of surface area A at an absolute temperature T is kept in a surrounding having a lower temperature \(T_{0}\). The net rate of loss of thermal energy from the body due to radiation is

\[\Delta u_{1} = e\sigma\text{ }A\left( {\text{ }T}^{4} - T_{0}^{4} \right)\]

If the temperature difference is small, we can write
or, \(\ T^{4} - T_{0} + \Delta T = \left( T_{0} + \Delta T \right)^{4} - T_{0}\ ^{4}\)

\[\begin{matrix} & \ = T_{0}^{4}\left( 1 + \frac{\Delta T}{{\text{ }T}_{0}} \right)^{4} - T_{0}^{4} \\ & \ = T_{0}^{4}\left\lbrack 1 + 4\frac{\Delta\text{ }T}{{\text{ }T}_{0}} + \text{~}\text{higher powers of}\text{~}\frac{\Delta T}{{\text{ }T}_{0}} \right\rbrack - T_{0}^{4} \\ & \ \approx 4{\text{ }T}_{0}^{3}\Delta\text{ }T = 4{\text{ }T}_{0}^{3}\left( \text{ }T - T_{0} \right) \end{matrix}\]

Thus, \(\ \Delta u_{1} = 4e\sigma{AT}_{0}\ ^{3}\left( \text{ }T - T_{0} \right)\)

\[= b_{1}\text{ }A\left( \text{ }T - T_{0} \right)\]

The body may also lose thermal energy due to convection in the surrounding air. For small temperature difference, the rate of loss of heat due to convection is also proportional to the temperature difference and the area of the surface. This rate may, therefore, be written as

\[\Delta u_{2} = b_{2}\text{ }A\left( \text{ }T - T_{0} \right)\]

The net rate of loss of thermal energy due to convection and radiation is

\[\Delta u = \Delta u_{1} + \Delta u_{2} = \left( b_{1} + b_{2} \right)A\left( \text{ }T - T_{0} \right).\]

If \(s\) be the specific heat capacity of the body and \(m\) its mass, the rate of fall of temperature is

\[\begin{matrix} & \frac{- dT}{dt} = \frac{\Delta u}{\text{ }ms} = \frac{b_{1} + b_{2}}{\text{ }ms}\text{ }A\left( \text{ }T - T_{0} \right) \\ & \ = bA\left( \text{ }T - T_{0} \right) \end{matrix}\]

Thus, for small temperature difference between a body and its surrounding, the rate of cooling of the body is directly proportional to the temperature difference and the surface area exposed. We can write

\[\frac{dT}{dt} = - bA\left( \text{ }T - T_{0} \right)\]

Cooling curve:

The law holds good only for small difference of temperature. Also, the loss of heat by radiation depends upon the nature of the surface of the body and the area of the exposed surface. We can write

\[- \frac{dT}{dt} = k\left( \text{ }T - T_{s} \right)\]

where \(k\) is a positive constant depending upon the area and nature of the surface of the body. Suppose a body of mass \(m\) and specific heat capacity \(s\) is at temperature \(T\). Let \(T_{s}\) and \(T_{o}\) be the temperature of the surroundings and body respectively. If the temperature falls by a small amount dT in time dt , then the amount of heat lost is

\[dQ = msdT\]

\(\therefore\ \) Rate of loss of heat is given by

\[\frac{dQ}{dt} = ms\frac{dT}{dt}\]

from equation
\(- \frac{dQ}{dt} = k\left( T - T_{s} \right)\) and \(\frac{dQ}{dt} = ms\frac{dT}{dt}\)
we have \(\ - ms\frac{dT}{dt} = k\left( T - T_{s} \right)\)

\[\frac{dT}{\text{ }T - T_{s}} = - \frac{k}{\text{ }ms}dt = - Kdt\ (\text{~}\text{where}\text{~}K = k/ms)\]

On integrating,

\[\begin{matrix} & \ln\left( \frac{T - T_{s}}{{\text{ }T}_{0} - T_{s}} \right) = - kt \\ & \text{ }T = T_{s} + \left( T_{0} - T_{s} \right)e^{- kt} \end{matrix}\]

enables you to calculate the time of colling of a body through a particular range of temperature.

Practice Exercise

Q. 1 A metal sphere with a black surface and radius 30 mm , is cooled to \(- 73C(200\text{ }K)\) and placed inside an enclosure at a temperature of \(27^{\circ}C(300\text{ }K)\). Calculate the initial rate of temperature rise of the sphere, assuming the sphere is a black body. (Assume density of metal \(= 8000\text{ }kg{\text{ }m}^{- 3}\) specific heat capacity of metal \(= 400\text{ }J{\text{ }kg}^{- 1}{\text{ }K}^{- 1}\), and Stefan constant \(= 5.7 \times 10^{- 8}\text{ }W{\text{ }m}^{- 2}{\text{ }K}^{- 4}\) ).
Q. 2 A pan filled with hot food cools from \(94^{\circ}C\) to \(86^{\circ}C\) in 2 minutes when the room temperature is at \(20^{\circ}C\). How long will it take to cool from \(71^{\circ}C\) to \(69^{\circ}C\) ?
Q. 3 A body cools from \(50^{\circ}C\) to \(40^{\circ}C\) in 5 minutes. The surrounding temperature is \(20^{\circ}C\). What will be its temperature 5 minutes after reading \(40^{\circ}C\) ? Use approximate method.

Answers

Q. \(1\ 0.012\text{ }K{\text{ }s}^{- 1}\) (approx).
Q. 242 sec
Q. \(3\ \text{ }T = \frac{100}{3}^{\circ}C\)

The wavelength corresponding to highest intensity m is inversely proportional to the absolute temperature. Thus
\[\lambda_{m} = \frac{b}{T} \]where \(b\left( = 2.89 \times 10^{- 3} \right.\ \) meter Kelvin) is known as the Wien's constant.
When the temperature of a black body is increased, the contribution of low wavelength radiation increases. This explains why a body on heating first appears red, then orange, then white and finally blue. This law also helps us in determining the temperatures of the stars.

Energy Distribution in Black Body Radiation:

The radiation emitted by a black body at any temperature is a mixture of all wavelengths. The graph shows qualitative variation in intensity wavelength, at different temperatures.

Spectral emmissive power:

To speak of the intensity of a single wavelength is meaningless. The slit of the spectrometer always gathers a band of wavelengths the narrower the slit the narrower the band-and we always speak of the intensity of a given band. We express it as follows :

\[\text{~}\text{energy radiated}\text{~}m^{- 2}{\text{ }s}^{- 1}\text{, in band}\text{~}\lambda\text{~}\text{to}\text{~}\lambda + \Delta\lambda = E_{\lambda}\Delta\lambda\]

The quantity \(E_{\lambda}\) is called emissive power of a black body for the wavelength \(\lambda\) and at the given temperature ; its definition follows from equation \(\lambda\) to \(\lambda + \Delta\lambda = E_{\lambda}\Delta\lambda\) :

\[\begin{matrix} & E_{\lambda} = \frac{\text{~}\text{energy radiated}\text{~}m^{- 2}{\text{ }s}^{- 1}\text{, in band}\text{~}\lambda\text{~}\text{to}\text{~}\lambda + \Delta\lambda}{\text{~}\text{band width,}\text{~}\Delta\lambda} \\ & E_{\lambda} = \frac{\text{~}\text{power radiated}\text{~}m^{- 2}\text{~}\text{in band}\text{~}\lambda\text{~}\text{to}\text{~}\lambda + \Delta\lambda}{\Delta\lambda} \end{matrix}\]

In the figure, \(E_{\lambda}\) is expressed in watts per \(m^{2}\) per nanometre \(\left( 10^{- 9}\text{ }m \right)\).
The quantity \(E_{\lambda}\Delta\lambda\) in equation \(\lambda\) to \(\lambda + \Delta\lambda = E_{\lambda}\Delta\lambda\) is the area beneath the radiation curve between the wavelength \(\lambda\) and \(\Delta\lambda\) (figure). Thus the energy radiated per meter \(\ ^{2}\) per second between those wavelengths in proportional to that area.
Similarly, the total radiation emitted per metre \(\ ^{2}\) per second over all wavelengths is proportional to the area under the whole curve.

Figure : Definition of \(E \cdot \lambda_{m}\) and \(E_{\lambda m}\)

Laws of black body radiation:

The curves of figure can be explained only Planck's quantum theory of radiation, which is outside our scope. Both theory and experiment lead to three generalisations, which together describe well the properties of black body radiation.
(i) If \(\lambda_{m}\) is the wavelength of the peak of the curve for T (in K ), then

\[\begin{array}{r} \lambda_{m}\text{ }T = constant\#(2) \end{array}\]

The value of the constant is \(2.9 \times 10^{- 3}\text{ }m\text{ }K\). In figure the dotted line is the locus of the peaks of the curves for different temperatures.

The relationship in (2) is sometimes called Wien's displacement law.
(ii) If \(E_{\lambda m}\) is the height of the peak of the curve for the temperature T , then

\[\begin{array}{r} E_{\lambda m} \propto {\text{ }T}^{5}\#(3) \end{array}\]

Figure : Distribution of intensity in black body radiation

(iii) If E is the total energy radiated per metre \(\ ^{2}\) per second at a temperature T , which is represented by the total area under the particular \(E_{\lambda} - \lambda\) curve, then

\[E = \sigma T^{4}\]

So in figure, which shows four \(E_{\lambda} - \lambda\) graphs at different temperatures T , the total area below the graphs should be proportional to the corresponding value of \(T^{4}\).

Q. 1 The diagram shows four slabs of different materials with equal thickness, placed side by side. Heat flows from left to right and the steady-state temperatures of the interfaces are given. Rank the materials according to their thermal conductivities, smallest to largest.

Sol In steady state heat transfered through each slab is same.
Than \(\frac{\Delta Q}{\Delta t} = \frac{k_{1}\text{ }A\Delta{\text{ }T}_{1}}{\text{ }d} = \frac{k_{2}\text{ }A\Delta{\text{ }T}_{2}}{\text{ }d} = \frac{k_{3}\text{ }A\Delta{\text{ }T}_{3}}{\text{ }d} = \frac{k_{4}\text{ }A\Delta{\text{ }T}_{4}}{\text{ }d}\)
\(k \propto \frac{1}{\Delta\text{ }T}\) thus \(3,4,2,1\)
Q. 2 (Growth of ice on Pond)

On a cold winter day, the atmospheric temperature is \(- \theta\) (on Celsius scale) which is below \(0^{\circ}C\). A cylindrical drum of height \(h\) made of a bad conductor is completely filled with water at \(0^{\circ}C\) and is kept outside without any lid. Calculate the time taken for the whole mass of water to freeze. Thermal conductivity of ice is K and its latent heat of fusion is L . Neglect expansion of water on freezing.

Sol.

Suppose, the ice starts forming at time \(t = 0\) and a thickness x is formed at time t . The amount of heat flown from the water to the surrounding in the time interval \(t\) to \(t = dt\) is

\[\Delta Q = \frac{KA\theta}{x}dt\]

The mass of the ice formed due to the loss of this amount of heat is

\[dm = \frac{\Delta Q}{\text{ }L} = \frac{KA\theta}{x}dt.\]

The thickness dx of ice formed in time dt is

\[dx = \frac{dm}{\text{ }A\rho} = \frac{K\theta}{\rho xL}dt\]

or, \(\ dt = \frac{\rho L}{K\theta}xdx\)
Thus, the time \(T\) taken for the whole mass of water to freeze is given by

\[\begin{matrix} & \ \int_{0}^{T}\mspace{2mu}\mspace{2mu} dt = \frac{\rho L}{\text{ }K\theta}\int_{0}^{h}\mspace{2mu}\mspace{2mu} xdx \\ & \text{ }T = \frac{\rho{Lh}^{2}}{2\text{ }K\theta} \end{matrix}\]

Q. 3 Two thin metallic spherical shells of radii \(r_{1}\) and \(r_{2}\left( r_{1} < r_{2} \right)\) are placed with their centres coinciding. A material of thermal conductivity K is filled in the space between the shells. The inner shell is maintained at temperature \(\theta_{1}\) and the outer shell at temperature \(\theta_{2}\left( \theta_{1} < \theta_{2} \right)\). Calculate the rate at which heat flows radially through the material.
Sol. Let us draw two spherical shells of radii x and \(x + dx\) concentric with the given system. Let the temperatures at these shells be \(\theta\) and \(\theta + d\theta\) respectively. The amount of heat flowing radially inward through the material between x and \(x + dx\) is

\[\frac{\Delta Q}{\Delta t} = \frac{K4\pi x^{2}d\theta}{dx}\]

Thus, \(\ K4\pi\int_{\theta_{1}}^{\theta_{2}}\mspace{2mu}\text{ }d\theta = \frac{\Delta Q}{\Delta t}\int_{\eta}^{r_{2}}\mspace{2mu}\frac{dx}{x^{2}}\)

\[\begin{matrix} & K4\pi\left( \theta_{2} - \theta_{1} \right) = \frac{\Delta Q}{\Delta t}\left( \frac{1}{r_{1}} - \frac{1}{r_{2}} \right) \\ & \frac{\Delta Q}{\Delta t} = \frac{4\pi{Kr}_{1}r_{2}\left( \theta_{2} - \theta_{1} \right)}{r_{2} - r_{1}} \end{matrix}\]

Q. 4 Three rods \(AB,BC\) and BD having thermal conductivities in the ratio \(1:2:3\) and lengths in the ratio \(2:1:1\) are joined as shown is figure. The ends \(A,C\) and \(D\) are at temperatures \(T_{1},T_{2}\) and \(T_{3}\) respectively. Find the temperature of the junction B. Assume steady state.

Sol. Let the thermal conductivities of the rods \(AB,BC\) and BD be \(K,2\text{ }K\) and 3 K respectively. Also, let their lengths be \(2\text{ }L,\text{ }L\) and L .
If T be the required temperature of the junction B and assuming \(T_{1} > T > T_{2} > T_{3}\), we have

\[\left. \ \left. \ \left. \ \frac{\Delta Q}{\Delta t} \right\rbrack_{AB} = \frac{\Delta Q}{\Delta t} \right\rbrack_{BC} + \frac{\Delta Q}{\Delta t} \right\rbrack_{BD}\]

i.e. \(\ \frac{KA\left( T_{1} - T \right)}{2\text{ }L} = \frac{2KA\left( T - T_{2} \right)}{L} + \frac{3KA\left( T - T_{3} \right)}{L}\)
or \(\ \frac{T_{1} - T}{2} = 2\left( \text{ }T - T_{2} \right) + 3\left( \text{ }T - T_{3} \right)\)
or \(\ T = \frac{1}{11}\left( {\text{ }T}_{1} + 4{\text{ }T}_{2} + 6{\text{ }T}_{3} \right)\)
Q. 5 The container A is constantly maintained at \(100^{\circ}C\) and insulated container B of the figure initially contains ice at \(0^{\circ}C\). Different rods are used to connect them. For a rod made of copper, it takes 30 minutes for the ice to melt and for a rod of steel of same cross-section taken in different experiment it takes 60 minutes for ice to melt. When these rods are simultaneously connected in parallel. Find the time interval in which ice melts?

Sol. \(\ Q = it\ \) where \(i =\) heat flow rate \(= \frac{\Delta T}{R} = \frac{100}{R}\)
For copper rod :

\[Q = \left( \frac{100}{R_{1}} \right)(30)\ \Rightarrow \ R_{1} = \left( \frac{100}{Q} \right) \times 30\]

Also for steel rod:

\[Q = \left( \frac{100}{R_{2}} \right) \times 60\ \Rightarrow \ R_{2} = \left( \frac{100}{Q} \right) \times 60\]

Now, \(Q = \left( \frac{1}{R_{1}} + \frac{1}{R_{2}} \right)t_{\text{req.}\text{~}}\)

\[\therefore\ t_{reqd} = \frac{Q}{\left( \frac{Q}{100} \right) \times \frac{1}{30} + \left( \frac{Q}{100} \right) \times \frac{1}{60}} = 20\text{ }\min\]

Q. 6 (Temperature of Sun)

Estimate the temperature \(T_{c}\) of the earth, assuming it is in radiative equilibrium with the sun. (Assume radius of sun, \(r_{s} = 7 \times 10^{8}\text{ }m\), temperature of solar surface \(= 6000\text{ }K\), distance of earth from sun, \(R = 1.5 \times 10^{11}\text{ }m\) )
Sol. Power radiated from sun \(= \sigma \times\) surface area \(\times T^{4}\)

\[= \sigma \times 4\pi r_{s}^{2} \times T_{s}^{4}\]

Power received by earth \(= \frac{\pi r_{c}^{2}}{4\pi R^{2}} \times\) power radiated by sun
since \(\pi r_{e}^{2}\) is the effective area of the earth on which the sun's radiation is incident normally. figure and \(4\pi R^{2}\) is the total area on which the sun's radiation falls at a distance \(R\) from the sun where the earth is situated.

\[\text{~}\text{Now power radiated by earth}\text{~} = \sigma \cdot 4\pi r_{e}^{2} \cdot T_{e}^{4}\]

Assuming radiative equilibrium
power radiated by earth \(=\) power received by earth

\[\therefore\ \sigma \cdot 4\pi r_{e}^{2} \cdot T_{e}^{4} = \sigma \cdot 4\pi r_{s}^{2} \cdot T_{s}^{4} \times \frac{\pi r_{e}^{2}}{4\pi R^{2}}\]

Cancelling \(r_{e}\ ^{2}\) and simplifying, then

\[T_{e}\ ^{4} = T_{s}\ ^{4} \times \left( \frac{r_{s}^{2}}{4R^{2}} \right)\]

\[\therefore\ T_{e} = T_{s} \times \left( \frac{7 \times 10^{8}}{2 \times 1.5 \times 10^{11}} \right)^{1/2} = 209\text{ }K \]Note that the calculation is approximate, for example, the earth and the sun are not perfect black body radiators and the earth receives heat from its interior.
Q. 7 A wood-burning stove stands unused in a room where the temperature is \(18^{\circ}C(291\text{ }K)\). A fire is started inside the stove. Eventually, the temperature of the stove surface reaches a constant \(198^{\circ}C(471\text{ }K)\), and the room warms to a constant \(29^{\circ}C(302\text{ }K)\). The stove has an emissivity of 0.900 and a surface area of \(3.50{\text{ }m}^{2}\). Determine the net radiant power generated by the stove when the stove (a) is unheated and has a temperature equal to room temperature and (b) has a temperature of \(198^{\circ}C\).

Sol. Reasoning: The stove emits more radiant power heated than when unheated. In both cases, however, the Stefan-Bolfzmann law can be used to determine the amount of power emitted. Power is energy per unit time or Q/t. But in this problem we need to find the net power produced by the stove. The net power is the power the stove emits minus the power the stove absorbs. Then power the stove absorbs comes from the wall, ceiling, and floor of the room, all of which emit radiation.
(a) Remembering that temperature must be expressed in kelvins when using the Stefan-Boltzmann law, we find that

Power emitted by unheated \(= \frac{Q}{t} = e\sigma^{T}A^{A}\)
\[= (0.900)\left\lbrack 5.67 \times 10^{- 8}\text{ }J/\left( s \cdot m^{2} \cdot {\text{ }K}^{4} \right\rbrack(291\text{ }K)^{4}\left( 3.50{\text{ }m}^{2} \right) = 1280\text{ }W \right.\ \]The fact that the unheated stove emits 1280 W of power and yet maintains a constant temperature means that the stove also absorbs 1280 W of radiant power from its surroundings. Thus, the net power generated by the unheated stove is zero.

Net power generated by stove at \(18^{\circ}C = \underset{\begin{matrix} \text{~}\text{Power emitted}\text{~} \\ \text{~}\text{by stove at}\text{~} \\ 18^{\circ}C \end{matrix}}{\overset{1280\text{ }W}{︸}} - \underset{\begin{matrix} \text{~}\text{Poocer}\text{ emitted by}\text{~} \\ \text{~}\text{room at}\text{~}18^{\circ}\text{~}\text{and}\text{~} \\ \text{~}\text{absorbed by stove}\text{~} \end{matrix}}{\overset{1280\text{ }W}{︸}} = 0\text{ }W\)
(b) The hot stove \(\left( 198^{\circ}C \right)\) or 471 K ) emits more radiant power than it absorbs from the cooler room. The radiant power the stove emits is
Power emitted by stove at \(198^{\circ}C = \frac{Q}{t} = e{\sigma T}^{4}\text{ }A\)
\[= (0.900)\left\lbrack 5.67 \times 10^{- 8}\text{ }J/\left( s \cdot m^{2} \cdot {\text{ }K}^{4} \right) \right\rbrack(471\text{ }K)^{4}\left( 3.50{\text{ }m}^{2} \right) = 8790\text{ }W \]The radiant power the stove absorbs from the room is identical to the power that the stove would emit at the constant room temperature of \(29^{\circ}C(302\text{ }K)\). The reasoning here is exactly like that in part (a).

Power emitted by
room at \(29^{\circ}C\) and \(\ = \frac{Q}{t} = e\sigma{\text{ }T}^{4}\text{ }A\)
absorbed by stove

\[= (0.900)\left\lbrack 5.67 \times 10^{- 8}\text{ }J/\left( s \cdot {\text{ }m}^{2} \cdot {\text{ }K}^{4} \right) \right\rbrack(302\text{ }K)^{4}\left( 3.50{\text{ }m}^{2} \right) = 1490\text{ }W\]

The net radiant power the stove produces from the fuel it burn is
Net power

\[\begin{matrix} & \text{~}\text{generated by}\text{~} \\ & \text{~}\text{stove at}\text{~}198^{\circ}C \end{matrix} = \underset{\begin{matrix} \text{~}\text{Power emitted}\text{~} \\ \text{~}\text{by stove at}\text{~} \\ 198^{\circ}C \end{matrix}}{\overset{8790\text{ }W}{︸}} - \underset{\begin{matrix} \text{~}\text{Power emitted by}\text{~} \\ \text{~}\text{room at 2}\text{~}9^{\circ}\text{~}\text{C}\text{~}\text{~}\text{and}\text{~} \\ \text{~}\text{absorbed by stove}\text{~} \end{matrix}}{\overset{1490\text{ }W}{︸}} = 7300\text{ }W\]

Q. 8 The room heater can provide only \(16^{\circ}C\) in the room when the temperature outside is \(- 20^{\circ}C\). It is not warm and comfortable, that is why the electric stove with power of 1 kW is also plugged in. Together these two devices maintain the room temperature of \(22^{\circ}C\). Determine the thermal power of the heater.

Sol. Rate of heat loss with only room heater
\(P_{h} = \frac{\Delta Q}{\Delta t} = C(16 + 20)\ \lbrack\) where \(C =\) constant \(\rbrack\)
while both heater and stove it is
\[{P_{h} + P_{s} = \left( \frac{\Delta Q}{\Delta t} \right)^{'} = C(22 + 20) }{\therefore\frac{P_{h}}{P_{h} + P_{s}} = \frac{36}{42} \Rightarrow 7P_{h} = 6P_{h} + 6P_{s} }{\Rightarrow P_{h} = 6P_{s} = 6\text{ }kW }\]Q. 9 A hot body palced in air is cooled down according to Newton's law of cooling, the rate of decrease of temperature being \(k\) times the temperature difference from the surrounding. Starting from \(t = 0\), find the time in which the body will loose half the maximum heat it can lose.
Sol. We have,

\[\frac{d\theta}{dt} = - k\left( \theta - \theta_{0} \right)\]

where \(\theta_{0}\) is the temperature of the surrounding and \(\theta\) is the temperature of the body at time \(t\). Suppose \(\theta = \theta_{1}\) at \(t = 0\)
Then,

\[\int_{\theta_{1}}^{\theta}\mspace{2mu}\frac{d\theta}{\theta - \theta_{0}} = - k\int_{0}^{t}\mspace{2mu} dt\]

or \(\ In\frac{\theta - \theta_{0}}{\theta_{1} - \theta_{0}} = - kt\)
or, \(\theta - \theta_{0} = \left( \theta_{1} - \theta_{0} \right)e^{- kt}\)

The body continues to lose heat till its temperature becomes equal to that of the surrounding. The loss of heat in this entire period is

\[\Delta Q_{m} = ms\left( \theta_{1} - \theta_{0} \right)\]

This is the maximum heat the body can looses. If the body loses half this heat, the decreases in its temperature will be,

\[\frac{\Delta Q_{m}}{2\text{ }ms} = \frac{\theta_{1} - \theta_{0}}{2}\]

If the body loses this heat in time \(t_{1}\), the temperature at \(t_{1}\) will be

\[\theta_{1} - \frac{\theta_{1} - \theta_{0}}{2} = \frac{\theta_{1} + \theta_{0}}{2}\]

Putting these values of time and temperature in (i),

\[\frac{\theta_{1} + \theta_{0}}{2} - \theta_{0} = \left( \theta_{1} - \theta_{0} \right)e^{- kt}\]

or,

\[e^{- kt} = \frac{1}{2}\]

or,

\[t_{1} = \frac{In2}{k}\]

Q. 10 Three identical metal rods \(A,B\) and C are placed end to end and a temperature difference is maintained between the free ends of \(A\) and \(C\). If the thermal conductivity of \(B\left( K_{B} \right)\) is twice that of \(C\left( K_{C} \right)\) and half that of \(A\left( K_{A} \right),\left( K_{A} = 49w/mK \right)\) calculate the effective thermal conductivity of the system?

Sol.

\[\begin{matrix} & \begin{matrix} K_{B} = K_{A}/2 \\ {\text{ }K}_{C} = K_{A}/4 \end{matrix} \\ & \frac{1}{R_{eq}} = \frac{L}{\text{ }A}\left( \frac{1}{{\text{ }K}_{A}} + \frac{1}{{\text{ }K}_{B}} + \frac{1}{{\text{ }K}_{C}} \right) = \frac{3\text{ }L}{\text{ }A}\left( \frac{1}{{\text{ }K}_{eff}} \right) \\ K_{eff} = \frac{3{\text{ }K}_{A}}{7} = 21w/mK & \end{matrix}\]