Question

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.

Answer 1

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;

$A=πr^2=π×0.04^2=5.0265×10^{-3}m^2$

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

$\sigma$ =$5.67×10^{-8}\frac{W}{m^2K^4}$

Hence heat Loss per unit time is;

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

$dQ=1.2535J$

Answer;

1.2535J