Question

A sphere of radius 6.0cm at 1200oC is suspended in a vacuum in an enclosure at 500oC. Find

the rate of loss of heat, assuming that it is a black body

Answer 1

The net radiation heat current ($H_{net}$) from a body at temperature T to its surroundings at temperature T_s depends on both temperatures, the surface area A (A = $4\pi r^2$ for a sphere of radius r), the emissivity e (e=1 for a black body) and the Stefan-Boltzmann constant ($\sigma = 5.6704\times10^{-8}\frac{W}{m^2 \cdot K^4}$ ). We proceed to find $H_{net}$ as:

$H_{net}=Ae\sigma(T^4-T^4_s) \\ H_{net}=(4\pi (0.06\,m)^2)(1)(5.6704\times10^{-8}\frac{W}{m^2 \cdot K^4})[ (1473\,K)^4-(773\,K)^4] \\ H_{net}=11160.496\,W \approxeq 11.1605\,kW$

In conclusion, we find that the rate of loss of heat with the surroundings is about 11.1605 kW.

Reference:

• Sears, F. W., & Zemansky, M. W. (1973). University physics.