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Question

Physics Question on thermal properties of matter

A black body, at a temperature of 227C227^{\circ} C, radiates heat at a rate of 20calm2s1.20 \,cal\, m ^{-2} s ^{-1} . When its temperature is raised to 727C727^{\circ} C, the heat radiated by it in calm2s1cal \,m ^{-2} s ^{-1} will be closest to

A

40

B

160

C

320

D

640

Answer

320

Explanation

Solution

The temperature of the black body is T1=227C=500KT _{1}=227^{\circ} C =500\, K \therefore Using Stefan's law, the rate of heat radiation per unit area per unit time is E1=σT420=σ(500)4E_{1}=\sigma T^{4} \Rightarrow 20=\sigma(500)^{4} σ=20(500)4\Rightarrow \sigma=\frac{20}{(500)^{4}} Now the temperature of the blackbody is raised to T2=727C=1000KT _{2}=727^{\circ} C =1000 \,K \therefore Rate of heat radiation per unit area E2=σT24E_{2}=\sigma T_{2}^{4} 20(500)4×(1000)4 \Rightarrow \frac{20}{(500)^{4}} \times(1000)^{4} =20×24=320calm2s1=20 \times 2^{4}=320\, cal\, m ^{-2} s ^{-1}