Answer to Question #248279 in Molecular Physics | Thermodynamics for Sohan

Question #248279

Determine the net heat transfer through the black body disc with radius 4cm and mass 9g, at 320C when transferred to a surrounding temperature of 650C.



1
Expert's answer
2021-10-10T16:49:23-0400

Solution;

The disk gains heat from the sorrounding,"Q_{in}" ,due to thermal conductivity and losses heat,"Q_{out}" ,due to radiation. But at steady state,the rate of heat loss is equal to rate of heat gain.

By Stefan-Boltzmann's law,the heat loss by radiation;

"\\frac{dQ_{out}}{dt}=A\\sigma(T_s^4-T_d^4)"

A is surface area of the disc;

"A=\u03c0r^2=\u03c0\u00d70.04^2=5.0265\u00d710^{-3}m^2"

"\\sigma" is the Stefan-Boltzmann's constant,take the value for a black body radiation as;

"\\sigma" ="5.67\u00d710^{-8}\\frac{W}{m^2K^4}"

Hence heat Loss per unit time is;

"dQ" =5.0265×10-3×5.67×10-8(3384-3054)

"dQ=1.2535J"

Answer;

1.2535J








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