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 equal to
A
(ϵQϵP)1/4θP
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B
(ϵPϵQ)1/4θP
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C
(ϵQϵP)1/4×1θP
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D
(ϵQϵP)4θP
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Solution
The correct option is B(ϵPϵQ)1/4θP Both P and Q have same radiant power. EPEQ=1 and both have same surface area. So, SPSQ=1 Now from Stefan-Boltzmann law EPEQ=ϵPSPθ4PϵQSQθ4Q θQ=(ϵPϵQ)1/4θP