Asked by robert
In one room the radiator has a dark color (emissitivity = 0.818). It has a temperature of 64.8 oC. The new owner of the house paints the radiator a lighter color (emissitivity = 0.417). 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?
Answers
Answered by
drwls
The Stefan-Boltzmann equation must use temperature in K.
T1 = 64.8 C = 338.0 K
If the radiating area stays the same,
0.818 T1^4 = 0.417 T2^4, so
T2 = 1.1835*T1 = 400.0 K = 126.8 C
T1 = 64.8 C = 338.0 K
If the radiating area stays the same,
0.818 T1^4 = 0.417 T2^4, so
T2 = 1.1835*T1 = 400.0 K = 126.8 C
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