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Radiation

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Special Distribution of Blackbody Radiation - Wien's Law

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Q: A body cools from \${80}^{\circ}\$C to \${50}^{\circ}\$C in \$5\$ minutes. Calculate the time it takes to cool from \${60}^{\circ}\$C to \${30}^{\circ}\$C. The temperature of the surroundings is \${20}^{\circ}\$C.

Here, Initial temperature \$( {T}_{i}) = {80}^{\circ}\$C Final temperature \$( {T}_{f}) = {50}^{\circ}\$C Temperature of the surrounding \$( {T}_{0}) = {20}^{\circ}\$C \$ t = 5\$ min According to Newton's law of cooling,Rate of cooling \$\dfrac{dT}{dt} = K[\dfrac{\left({T}_{i}+{T}_{f}\right)}{2}-{T}_{o}]\$\$\dfrac{\left({T}_{f}-{T}_{i}\right)}{t}=K\left[\dfrac{\left(80 +50\right)}{2} - 20\right]\$\$\dfrac{80-50}{5} = K[ 65 - 20]\$\$6 = K\times 45\$\$K = \dfrac{6}{45} = \dfrac{2}{15}\$ In second condition, initial temperature \$={T}_{i}= {60}^{\circ}\$C Final temperature \${T}_{f}={30}^{\circ}\$C Time taken for cooling is \$t\$ According to Newton's law of cooling \$\dfrac{( 60 - 30)}{t} = \dfrac{2}{15} \left[ \dfrac{\left(60+30\right)}{2} -20\right]\$\$\dfrac{30}{t}=\dfrac{2}{15}\times 25\$\$\dfrac{30}{t} = \dfrac{50}{15}=\dfrac{10}{3}\$\$\therefore t = 9\$ min

Q: A spherical black body with a radius of 12cm radiates 450W power at 500K. If the radius were halved and the temperature doubled, the power radiated in watts would be :

We know that radiation from the body is given as \$E=\epsilon \sigma A{ T }^{ 4 }\$\$\dfrac { { E }_{ 1 } }{ { E }_{ 2 } } =\dfrac { { r }_{ 1 }^{ 2 }{ \times T }_{ 1 }^{ 4 } }{ { r }_{ 2 }^{ 2 }{ \times T }_{ 2 }^{ 4 } } =\dfrac { { 0.12 }^{ 2 }{ \times 500 }^{ 4 } }{ { 0.06 }^{ 2 }\times { 1000 }^{ 4 } } \$\${ E }_{ 2 }=\dfrac { 450\times { 0.06 }^{ 2 }\times { 1000 }^{ 4 } }{ { 0.12 }^{ 2 }\times { 500 }^{ 4 } } =1800\ watt\$

Q: The temperature of equal masses of three different liquids \$A, B\$ and \$C\$ are \$12^{\circ}C, 19^{\circ}C\$ and \$28^{\circ}C\$ respectively. The temperature when \$A\$ and \$B\$ are mixed is \$16^{\circ}C\$ and when \$B\$ and \$C\$ are mixed it is \$23^{\circ}C\$. What should be the temperature when \$A\$ and \$C\$ are mixed?

Let \$m\$ be the mass of each liquid and \$S_{A}, S_{B}, S_{C}\$ be specific heats of liquids \$A, B\$ and \$C\$ respectively. When \$A\$ and \$B\$ are mixed. The final temperature is \$16^{\circ}C\$.\$\therefore\$ Heat gained by \$A =\$ heat lost by \$B\$i.e., \$mS_{A} = (16 - 12) = mS_{B} (19 - 16)\$i.e., \$S_{B} = \dfrac {4}{3}S_{A} ..... (i)\$When \$B\$ and \$C\$ are mixed. Heat gained by \$B =\$ Heat lost by \$C\$i.e., \$mS_{B} = (23 - 19) = mS_{C} (28 - 23)\$i.e., \$S_{C} = \dfrac {4}{5}S_{B} ..... (ii)\$From eq. (i) and (ii)\$S_{C} = \dfrac {4}{5}\times \dfrac {4}{3} S_{A} = \dfrac {16}{15}S_{A}\$When \$A\$ and \$C\$ are mixed, let the final temperature be \$\theta\$\$\therefore mS_{A} (\theta - 12) = mS_{C} (28 - \theta)\$i.e., \$\theta - 12 = \dfrac {16}{15} (28 - \theta)\$By solving, we get,\$\theta = \dfrac {628}{31} = 20.26^{\circ}C\$.

Q: A black body at a temperature of \$227^{\circ}C\$ radiates heat energy at the rate \$5\;cal/cm^2/s\$. At a temperature of \$727^{\circ}C\$, the rate of heat radiated per unit area in \$cal/cm^2\$ will be:

Heat energy per unit time \$R = \sigma AeT^4\$Thus heat energy per unit area \$R' = \sigma e T^4\$\$\implies\$ \$\dfrac{R'_2}{R'_1} = \bigg( \dfrac{T_2}{T_1} \bigg)^4\$Initial temperature of the body \$T_1 =227 ^oC = (227+ 273) K = 500 K\$Final temperature of the body \$T_2 =727 ^oC = (727+ 273) K = 1000 K\$\$\therefore\$ \$\dfrac{R'_2}{5} = \bigg( \dfrac{1000}{500} \bigg)^4\$\$\dfrac{R'_2}{5} = 16\$ \$\implies R'_2 = 80\$ \$cal/cm^2\$

Q: A pan filled with hot food cools from \$94^\circ C\$ to \$86^\circ C\$ in \$2\$ minutes. When the room temperature is \$20^\circ C\$. How long will it cool from \$74^\circ C\$ to \$66^\circ C\$?

Using Newton's law of cooling with approximation, \$\dfrac{dT}{dt} = \dfrac{-k}{ms} (T_{avg} - T_s)\$Case 1 : \$dt = 2\$ minutes; \$T_s = 20\$ \$T_{avg} =\dfrac{94+86}{2} = 90\$ and \$dT = 94-86 = 8\$\$\therefore\$ \$\dfrac{-k}{ms} = \dfrac{dT}{dt}.\dfrac{1}{T_{avg}-T_s} = \dfrac{8}{2\times 70} = 0.05714\$ .......(a)Case 2 : \$T_s = 20\$ \$T_{avg} =\dfrac{74+66}{2} = 70\$ and \$dT = 74-66 = 8\$\$\therefore\$ Time in which it cools down \$\dfrac{8}{dt} = 0.05714\times (70-20)\$\$\implies \$ \$dt = 2.8\$ minutes

Q: A body cools down from \$80^o C\$ to \$60^oC\$ in \$10\$ Minutes when the temperature of surrounding is \$30^oC\$. The temperature of the body after next \$10\$ minutes will be:

\$\cfrac{dT}{dt}=-K\left(T-T_s\right)+T_in\$\$\therefore\$ Here from newtons law of co0ling\$\cfrac{dT}{dt}=-k\left(\cfrac{T_f+T_i}{2}-T_s\right)\$\$\therefore \cfrac{T_f-T_i}{\triangle t}=K\left[\cfrac{T_f+T_i}{2}-T_s\right]\$\$\therefore K=\cfrac{1}{20}\$\$\Rightarrow=\cfrac{60-T}{10}=\cfrac{1}{20}\left[\cfrac{60+T}{2}-30\right]\$\$=120-2T=\cfrac{T+60}{21}-30\$\$\cfrac{5T}{2}=120\$\$T=48\$

Q: State Kirchhoff's law of radiation and prove it theoretically.

Kirchhoff's law of radiation : At a given temperature the coefficient of absorption of a body is equal to its coefficient of emission.Theoretical proof : Consider the following thought experiment. An ordinary body \$A\$ and a perfectly black body \$B\$ are enclosed in an athermanous enclosure as shown in figure.According to the theory of heat exchanges there will be a continuous exchange of heat energy between each body and its surroundings. Hence, the two bodies, after some time, will attain the same temperature as that of enclosure. Let \$a\$ and \$e\$ be the coefficients of absorption and emission respectively of body \$A\$ and body \$B\$.Let \$E\$ and \$E_{b}\$ be the emissive powers of bodies \$A\$ and \$B\$ respectively.Suppose that \$Q\$ is the quantity of radiant energy incident on each body per unit time per unit surface area of the body.Body \$A\$ will absorb the quantity a \$Q\$ per unit time per unit surface area and radiate the quantity \$E\$ per unit time per unit surface area.Since there is no change in its temperature, we must have\$a\ Q = E ... (1)\$As body \$B\$ is a perfect blackbody it will absorb the quantity \$Q\$ per unit time per unit surface area and radiate the quantity \$E_{b}\$ per unit surface area.Since there is no change in its temperature. We must have\$Q = E_{b} .... (2)\$From Equation \$(1)\$ and \$(2)\$ we get\$a = \dfrac {E}{Q} = \dfrac {E}{E_{b}} .... (3)\$By definition of coefficient of emission\$e = \dfrac {E}{E_{b}} .... (4)\$From Equation \$(3)\$ and \$(4)\$ we get \$a = e\$.Hence, Kirchhoff's law of radiation proved.

Q: State:Wien's displacement law

Wien's displacement law states that the black-body radiation curve for different temperatures will peak at different wavelengths that are inversely proportional to the temperature.

Q: Certain quantity of water cools from \$70^oC\$ to \$60^oC\$ in the first \$5\$ minutes and to \$54^oC\$ in the next \$5\$ minutes. The temperature of the surroundings is ;

\$\dfrac {60-70}{5}=-K(65-T)\$\$\dfrac {54-60}{5}-K(57-T)\$\$\dfrac {-10}{-6}=\dfrac {65-T}{57-T}\$\$285-5T=195-3T\$\$90=2T\$\$T=45^o\$

Q: A sphere , a cube and a thin circular plate,all of same materials and same mass, are initially heated to same high temperature.

The circular plate has maximum surface area and as such it cools the fastest. For a given volume (mass in this case),a sphere has the least surface area and accordingly cools the slowest.

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A sphere , a cube and a thin circular plate,all of same materials and same mass, are initially heated to same high temperature.

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Revise with Concepts

What is Radiation?

Stefan Boltzmann Law

Special Distribution of Blackbody Radiation - Wien's Law

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Quick Summary With Stories

Radiation

Introduction to Stefan Boltzmann law

Wien's Displacement Law

Working Principle Thermos Flask

A body cools from 80∘C to 50∘C in 5 minutes. Calculate the time it takes to cool from 60∘C to 30∘C. The temperature of the surroundings is 20∘C.

A spherical black body with a radius of 12cm radiates 450W power at 500K. If the radius were halved and the temperature doubled, the power radiated in watts would be :

The temperature of equal masses of three different liquids A,B and C are 12∘C,19∘C and 28∘C respectively. The temperature when A and B are mixed is 16∘C and when B and C are mixed it is 23∘C. What should be the temperature when A and C are mixed?

A black body at a temperature of 227∘C radiates heat energy at the rate 5cal/cm2/s. At a temperature of 727∘C, the rate of heat radiated per unit area in cal/cm2 will be:

A pan filled with hot food cools from 94∘C to 86∘C in 2 minutes. When the room temperature is 20∘C. How long will it cool from 74∘C to 66∘C?

A body cools down from 80oC to 60oC in 10 Minutes when the temperature of surrounding is 30oC. The temperature of the body after next 10 minutes will be:

State Kirchhoff's law of radiation and prove it theoretically.

State: Wien's displacement law

Certain quantity of water cools from 70oC to 60oC in the first 5 minutes and to 54oC in the next 5 minutes. The temperature of the surroundings is ;

A sphere , a cube and a thin circular plate,all of same materials and same mass, are initially heated to same high temperature.

A body cools from $80^{\circ}$ C to $50^{\circ}$ C in $5$ minutes. Calculate the time it takes to cool from $60^{\circ}$ C to $30^{\circ}$ C. The temperature of the surroundings is $20^{\circ}$ C.

A black body at a temperature of $22 7^{\circ} C$ radiates heat energy at the rate $5 c a l / c m^{2} / s$ . At a temperature of $72 7^{\circ} C$ , the rate of heat radiated per unit area in $c a l / c m^{2}$ will be:

A pan filled with hot food cools from $9 4^{\circ} C$ to $8 6^{\circ} C$ in $2$ minutes. When the room temperature is $2 0^{\circ} C$ . How long will it cool from $7 4^{\circ} C$ to $6 6^{\circ} C$ ?

A body cools down from $8 0^{o} C$ to $6 0^{o} C$ in $10$ Minutes when the temperature of surrounding is $3 0^{o} C$ . The temperature of the body after next $10$ minutes will be:

State:
Wien's displacement law

Certain quantity of water cools from $7 0^{o} C$ to $6 0^{o} C$ in the first $5$ minutes and to $5 4^{o} C$ in the next $5$ minutes. The temperature of the surroundings is ;