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Question: Two bodies \(P\) and \(Q\) have thermal emissivities of \({\varepsilon _P}\), \({\varepsilon _Q}\) r...

Two bodies PP and QQ have thermal emissivities of εP{\varepsilon _P}, εQ{\varepsilon _Q} respectively. Surface areas of these bodies are the same and the total radiant power is also emitted at the same rate. If the temperature of PP in Kelvin is θP{\theta _P} , then the temperature of QQ i.e. θQ{\theta _Q} in kelvin is –
A. (εQεP)1/4θP{\left( {\dfrac{{{\varepsilon _Q}}}{{{\varepsilon _P}}}} \right)^{1/4}}{\theta _P}
B. (εPεQ)1/4θP{\left( {\dfrac{{{\varepsilon _P}}}{{{\varepsilon _Q}}}} \right)^{1/4}}{\theta _P}
C. (εQεP)1/4×1θP{\left( {\dfrac{{{\varepsilon _Q}}}{{{\varepsilon _P}}}} \right)^{1/4}} \times \dfrac{1}{{{\theta _P}}}
D. (εQεP)4θP{\left( {\dfrac{{{\varepsilon _Q}}}{{{\varepsilon _P}}}} \right)^4}{\theta _P}

Explanation

Solution

We may use the concept of emissive power of the body which states that the quantity of radiant energy emitted by the body per unit time per unit surface area of the body at that temperature. Also, Stefan’s law of radiation which is the amount of radiant energy per unit time per unit surface area of a body is proportional to the fourth power of its absolute temperature.

Formulae used:
Stefan’s law of radiation
εαθ4\varepsilon \alpha {\theta ^4}
Where, ε\varepsilon - emissive power of the body at temperature θ\theta

Complete step by step answer:
Let us understand the given data in question,
EP,EQ{E_P},{E_Q}- radiant power of body P,QP,Q
SP,SQ{S_P},{S_Q} - surface area of the body P,QP,Q
θP,θQ{\theta _P},{\theta _Q} - temperatures of the body P,QP,Q , respectively.
We are given that bodies P,QP,Q have the same emissive power.
EP=EQ−−−−−−−−−−(1){E_P} = {E_Q} - - - - - - - - - - (1)
Also, the bodies P,QP,Q have same surface areas, then
SP=SQ−−−−−−−−−−(2){S_P} = {S_Q} - - - - - - - - - - (2)
Now, Using Stefan’s law of radiation, we have
εPεQ=EPSPθP4EQSQθQ4\dfrac{{{\varepsilon _P}}}{{{\varepsilon _Q}}} = \dfrac{{{E_P}{S_P}\theta _P^4}}{{{E_Q}{S_Q}\theta _Q^4}}
Using eq(1)&(2)(1)\& (2), we get
∴θQ=(εPεQ)1/4θP\therefore {\theta _Q} = {\left( {\dfrac{{{\varepsilon _P}}}{{{\varepsilon _Q}}}} \right)^{1/4}}{\theta _P}

Hence, option B is correct.

Note: We have substituted emissive power of a body into Stefan’s law of radiation.These laws are applied to perfectly black bodies. Every body radiates energy at all temperatures except at absolute zero temperature and the quantity of radiant energy emitted by the body per unit time per unit surface area of the body at that temperature.