posted by StressedStudent on .
In an old house, the heating system uses radiators, which are hollow metal devices through which hot water or steam circulates. In one room the radiator has a dark color (emissitivity = 0.804). It has a temperature of 65.9 oC. The new owner of the house paints the radiator a lighter color (emissitivity = 0.439). Assuming that it emits the same radiant power as it did before being painted, what is the temperature (in degrees Celsius) of the newly painted radiator?
(emissivity)*sigma*T^4 = constant
sigma is the Stefan-Bolzmann constant, which cancels out.
T must be in Kelvin
(0.804)*sigma*339.1^4 = (0.439)*sigma*T^4
(T/339.1)^4 = 1.8314
T/339.1 = 1.163
T = 394.5 K = 121 C