Question

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 = 0.820). It has a temperature of 60.4 oC. The new owner of the house paints the radiator a lighter color (emissitivity = 0.491). 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?

Answer 1

Radiant power, emitted by the body with area $A$ , temperature $T$ and emissivity $\epsilon$ is

$P = Aj= \epsilon \sigma T^4 A$ , where $j= \epsilon \sigma T^4$ is Stefan-Bolzmann law.

According to the task, $P_1 = P_2$

$\epsilon_1 \sigma T_1^4 A = \epsilon_2 \sigma T_2^4 A$

$\epsilon_1 T_1^4 = \epsilon_2 T_2^4$

Converting from Celsius to Kelvin,

$T_1 = t _1+ 273.15 = 60.4 + 273.15 = 333.55\, K$

$\displaystyle T_2 = (\frac{\epsilon_1}{\epsilon_2})^{\frac{1}{4}} T_1= (\frac{0.820}{0.491})^{0.25} 333.55 =1.1368 \cdot 333.55 =379.18 \; K$

$t_2 = T_1 - 273.15 = 106.03^\circ C$

Answer: $t_2 = 106.03^\circ C$