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>> $Kirchoff's Law of Heat Radi...$

Kirchoff's Law of Heat Radiation and its Theoretical Proof

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: 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: 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: 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: The coefficient of performance of a refrigerator is 5 .If the temperarture inside freezer is - 20\$^{0}\$C, the temperature of the surroundings to which it rejects heat is

\$5 = \dfrac{{heat}}{{work\,done}}\$\$5 = \dfrac{{T\left( {fr} \right)}}{{T\left( {surr} \right) - T\left( {fr} \right)}}\$\$5 = \dfrac{{253K}}{{T - 253}}\$\$5T - 1265 = 253\$\$5T = 1518\$\$T = 303.6K\,or\,{30.6^0}C\$We get, \$T = \,{30.6^0}C.\$

Q: A black body is at a temperature of \$5760 K\$. The energy of radiation emitted by the body at wavelength \$250 nm\$ is \${U}_{1}\$, at wavelength \$500 nm\$ is \${U}_{2}\$ and that at \$1000 nm\$ is \${U}_{3}\$. Wien's constant, \$b = 2.88 \times {10}^{6} nmK\$. Which of the following is correct?

Maximum amount of emitted radiation corresponding to \$\lambda_m=\dfrac{b}{T}\$\$\Rightarrow \lambda_m=\dfrac{2.88\times 10^6nmK}{5760K}=500\ nm\$From the graph we can see that \$U_1 U_3\$

Q: The plots of intensity versus wavelength for three black bodies at temperatures \$T_{1},T_{2}\$ and \$ T_{3}\$ respectively are shown in fig. Their temperatures are such that:

As we know that from emissive power \${E}{\propto}{T}^{4}\$ and from wein's displacement law \${\lambda}_{max}\times{T}=constant \$. So if emissivity will be maximum them simultaneously it's temperature will be maximum, but from the wein's law we can say that the particle which has maximum temperature will have lower wavelength. So, \${E}{\propto}{T}{\propto}\dfrac{1}{\lambda}\$So, we can say:\${T}_{1}>{T}_{3}>{T}_{2}\$\${E}_{1}>{E}_{3}>{E}_{2}\$But, \${\lambda}_{1}<{\lambda}_{3}<{\lambda}_{2}\$

Physics

law

Kirchoff's Law of Radiation

At a given temperature, the ratio of the emissive power of a body to its absorptive power is constant and is equal to the emissive power of a black body at the same temperature.

\frac{E}{a} = E_{b}

But

\frac{E}{E _{b}} = e

∴ a = e

Alternative statement of Kirchhoff's law: At any given temperature, the emissivity of a body is equal to its coefficient of absorption.

example

Problems using Kirchoff's Law

At a particular temperature and wavelength, the spectral emissive power and monochromatic absorptive power of a body are 10 and 8 units respectively. The emissive power of a black body at the same temperature will be:
According to Kirchoff's law.

\frac{e _{λ}}{a _{λ}} = E_{λ}

here, we are given that

e_{λ}

= 10 and

a_{λ}

= 8
So,

E_{λ}

\frac{1 0}{8}

= 1.25

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Stefan Boltzmann Law

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Special Distribution of Blackbody Radiation - Wien's Law

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Kirchhoff's law Part-I

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Kirchhoff's law Part II

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Quick Summary With Stories

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Mathematical explanation of Kirchhoff's law

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Important Questions

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 :

Medium

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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.

Medium

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The temperature of equal masses of three different liquids $A, B$ and $C$ are $1 2^{\circ} C, 1 9^{\circ} C$ and $2 8^{\circ} C$ respectively. The temperature when $A$ and $B$ are mixed is $1 6^{\circ} C$ and when $B$ and $C$ are mixed it is $2 3^{\circ} C$ . What should be the temperature when $A$ and $C$ are mixed?

Hard

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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:

Medium

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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 $6 6^{\circ} C$ ?

Medium

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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 ;

Medium

NEET

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A body cools from $6 0^{\circ} C$ to $5 0^{\circ} C$ in $10$ minutes. If the room temperature is $2 5^{\circ} C$ and assuming Newton's cooling law holds good, the temperature of the body at the end of the next $10$ minutes is:

Medium

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The coefficient of performance of a refrigerator is 5 .If the temperarture inside freezer is - 20 $^{0}$ C, the temperature of the surroundings to which it rejects heat is

Easy

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A black body is at a temperature of $5760 K$ . The energy of radiation emitted by the body at wavelength $250 n m$ is $U_{1}$ , at wavelength $500 n m$ is $U_{2}$ and that at $1000 n m$ is $U_{3}$ . Wien's constant, $b = 2.88 \times 10^{6} n m K$ . Which of the following is correct?

Medium

NEET

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The plots of intensity versus wavelength for three black bodies at temperatures $T_{1}, T_{2}$ and $T_{3}$ respectively are shown in fig. Their temperatures are such that:

Medium

View solution

law

Kirchoff's Law of Radiation

example

Problems using Kirchoff's Law

REVISE WITH CONCEPTS

What is Radiation?

Stefan Boltzmann Law

Special Distribution of Blackbody Radiation - Wien's Law

Quick Summary With Stories

Kirchhoff's law

Mathematical explanation of Kirchhoff's law

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 :

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.

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?

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 body cools from 60∘C to 50∘C in 10 minutes. If the room temperature is 25∘C and assuming Newton's cooling law holds good, the temperature of the body at the end of the next 10 minutes is:

The coefficient of performance of a refrigerator is 5 .If the temperarture inside freezer is - 200C, the temperature of the surroundings to which it rejects heat is

A black body is at a temperature of 5760K. The energy of radiation emitted by the body at wavelength 250nm is U1​, at wavelength 500nm is U2​ and that at 1000nm is U3​. Wien's constant, b=2.88×106nmK. Which of the following is correct?

The plots of intensity versus wavelength for three black bodies at temperatures T1​,T2​ and T3​ respectively are shown in fig. Their temperatures are such that:

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 $6 6^{\circ} C$ ?

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 ;

A body cools from $6 0^{\circ} C$ to $5 0^{\circ} C$ in $10$ minutes. If the room temperature is $2 5^{\circ} C$ and assuming Newton's cooling law holds good, the temperature of the body at the end of the next $10$ minutes is:

The coefficient of performance of a refrigerator is 5 .If the temperarture inside freezer is - 20 $^{0}$ C, the temperature of the surroundings to which it rejects heat is

A black body is at a temperature of $5760 K$ . The energy of radiation emitted by the body at wavelength $250 n m$ is $U_{1}$ , at wavelength $500 n m$ is $U_{2}$ and that at $1000 n m$ is $U_{3}$ . Wien's constant, $b = 2.88 \times 10^{6} n m K$ . Which of the following is correct?

The plots of intensity versus wavelength for three black bodies at temperatures $T_{1}, T_{2}$ and $T_{3}$ respectively are shown in fig. Their temperatures are such that: