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NEET PHYSICSEasy

A black body is at 727C727^\circ\text{C}. The rate at which it emits energy is proportional to:

A

(727)2(727)^2

B

(1000)4(1000)^4

C

(1000)2(1000)^2

D

(727)4(727)^4

Step-by-Step Solution

According to the Stefan-Boltzmann law, the rate at which a black body emits radiant energy is directly proportional to the fourth power of its absolute temperature (ET4E \propto T^4). Given the temperature of the black body, TC=727CT_C = 727^\circ\text{C}. Converting this to absolute temperature (Kelvin scale): T=TC+273=727+273=1000 KT = T_C + 273 = 727 + 273 = 1000 \text{ K}. Therefore, the rate at which it emits energy is proportional to T4=(1000)4T^4 = (1000)^4.

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