Two bodies P and Q have thermal emissivities of ϵP and ϵQ respectively. Surface areas of these bodies are same and the total radiant power is also emitted at the same rate. If temperature of P is θP kelvin then temperature of Q i.e. θQ is :
A
(ϵQϵP)14θP
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B
(ϵPϵQ)14θP
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C
(ϵQϵP)14×1θP
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D
(ϵQϵP)4θP
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Solution
The correct option is B(ϵPϵQ)14θP
EPEQ=ϵPSPθ4PϵQSQθ4Q (given EPEQ=1;SPSQ=1)
⇒θQ=(ϵPϵQ)14×θP
The ratio of temperatures will be inversely proportional to 1/4th power of their emmissivity as the surface area ratio and emitted energy ratios are 1.