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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 Stefan-Boltzmann Law, the energy radiated per unit area per unit time is directly proportional to the fourth power of the absolute temperature of the black body. ET4E \propto T^4 E2E1=(T2T1)4\frac{E_2}{E_1} = \left(\frac{T_2}{T_1}\right)^4 Given: E1=7 cal cm2s1E_1 = 7\text{ cal cm}^{-2}\text{s}^{-1} T1=227C=227+273=500 KT_1 = 227^{\circ}\text{C} = 227 + 273 = 500\text{ K} T2=727C=727+273=1000 KT_2 = 727^{\circ}\text{C} = 727 + 273 = 1000\text{ K} Substituting the values: E27=(1000500)4=24=16\frac{E_2}{7} = \left(\frac{1000}{500}\right)^4 = 2^4 = 16 E2=16×7=112 cal cm2s1E_2 = 16 \times 7 = 112\text{ cal cm}^{-2}\text{s}^{-1}

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