Thermal Properties of Matter

JEE Main level, numerical answer, medium

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Question 1 of 126

A black body at temperature T1=300 KT_1 = 300\,\text{K} radiates power 81 W81\,\text{W}. If its temperature is raised to T2=600 KT_2 = 600\,\text{K}, the radiated power is (in W):

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Answer: 1296

P∝T4  ⟹  P2P1=(600300)4=24=16  ⟹  P2=16×81=1296 WP \propto T^4 \implies \frac{P_2}{P_1} = \left(\frac{600}{300}\right)^4 = 2^4 = 16 \implies P_2 = 16 \times 81 = 1296\,\text{W}.

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