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NEET PHYSICSMedium

A black body at 227C227^\circ\text{C} radiates heat at the rate of 7 cal cm2s17 \text{ cal cm}^{-2}\text{s}^{-1}. At a temperature of 727C727^\circ\text{C}, the rate of heat radiated in the same units will be:

A

6060

B

5050

C

112112

D

8080

Step-by-Step Solution

According to the Stefan-Boltzmann law, the rate of heat radiated per unit area by a black body is directly proportional to the fourth power of its absolute temperature, i.e., ET4E \propto T^4. Given: Initial temperature, T1=227C=227+273=500 KT_1 = 227^\circ\text{C} = 227 + 273 = 500 \text{ K} Initial rate of heat radiated, E1=7 cal cm2s1E_1 = 7 \text{ cal cm}^{-2}\text{s}^{-1} Final temperature, T2=727C=727+273=1000 KT_2 = 727^\circ\text{C} = 727 + 273 = 1000 \text{ K} Therefore, the ratio of the new rate of heat radiated (E2E_2) to the initial rate (E1E_1) is: E2E1=(T2T1)4=(1000500)4=24=16\frac{E_2}{E_1} = \left(\frac{T_2}{T_1}\right)^4 = \left(\frac{1000}{500}\right)^4 = 2^4 = 16 E2=16×E1=16×7=112 cal cm2s1E_2 = 16 \times E_1 = 16 \times 7 = 112 \text{ cal cm}^{-2}\text{s}^{-1}

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