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

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Question

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

Sum
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Solution

Given: 

T1 = 827°C = 827 + 273 K = 1100 K

T2 = 427°C = 427 + 273 K = 700 K

T0 = 27°C = 27 + 273 K = 300 K

To find: R1 : R2 = ?

Formula:

`"R"_1 = (("dQ")/("dt"))_1 = "e"sigma"A"("T"_1^4 - "T"_0^4)`

`"R"_2 = (("dQ")/("dt"))_2 = "e"sigma"A"("T"_2^4 - "T"_0^4)`

`"R"_1/"R"_2 = ("T"_1^4 - "T"_0^4)/("T"_2^4 - "T"_0^4)`

∴ `"R"_1/"R"_2 = (1100^4 - 300^4)/(700^4 - 300^4) = 14560/2320 = 182/29`

∴ R1 : R2 = 182 : 29

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Stefan-boltzmann Law of Radiation
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