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 = 0.807). It has a temperature of 61.3 oC. The new owner of the house paints the radiator a lighter color (emissitivity = 0.497). 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?
According to Stefan-Boltzmann law
P = ε∙σ∙A∙T^4
(ε emissivity, σ Stefan-Boltzmann constant, A surface area)
Radiant power is the same for both paints
εlight *σ * A * (Tlight)^4 = εdark * σ ∙*A * (Tdark)^4
εlight * (Tlight)^4 = εdark *(Tdark)^4
Tlight = Tdark ∙ (εdark/εlight)^(1/4)
Tlight = (61.3+273)(0.807/0.497)^1/4 = 377.36 K = 104.36 C
In an old house, the heating system uses radiators, which are hollow metal devices through which...