Answer to Question #134378 in Mechanics | Relativity for Jessica

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: $t_2 = 106.03^\circ C$