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Question 81 of 90

Q.Compare the rate of loss of heat from a metal sphere at 827°C with rate of loss of heat from the same at 427°C, if the temperature of surrounding is 27°C.

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2023Subjective· 2mImportance★★★★★
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Rate of loss of heat by radiation follows Stefan's law with a surrounding-temperature correction.

By Stefan's law (with surroundings at temperature T0T_0), the net rate of loss of heat is R∝(T4−T04)R \propto (T^4-T_0^4).

Convert to kelvin: T1=827+273=1100T_1=827+273=1100 K, T2=427+273=700T_2=427+273=700 K, T0=27+273=300T_0=27+273=300 K.

R1R2=T14−T04T24−T04=(1100)4−(300)4(700)4−(300)4\frac{R_1}{R_2} = \frac{T_1^4-T_0^4}{T_2^4-T_0^4} = \frac{(1100)^4-(300)^4}{(700)^4-(300)^4}

Computing: 11004=1.4641×10121100^4 = 1.4641\times10^{12}, 7004=2.401×1011700^4=2.401\times10^{11}, 3004=8.1×109300^4=8.1\times10^{9}. …

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