Question

The shell of a space station is a blackened sphere in which a temperature $T = 500K$ is maintained due to the operation of appliances of the station. The amount of heat given away from a unit surface, the area is proportional to the fourth power of thermodynamic temperature.Determine the temperature ${T_x}$ of the shell if the station is enveloped by a thin spherical black screen of nearly the same radius as the radius of the shell.

Answer 1

In the given question, the space station has maintained the temperature of $500K$. If the space station is covered with a thin spherical black screen of nearly the same radius as the radius of the shell then the temperature ${T_x}$ of the envelope will be a little higher and can be calculated as given below.

Complete step by step answer:
The total amount of heat Q released into the room per unit time remains unchanged because it is determined by the energy released during the operation of the station equipment. Since only the outer surface of the screen emits into space (this radiation depends only on its temperature), the temperature of the screen must be equal to the initial temperature of the station $T = 500K$.

However, the screen emits the same amount of heat Q inward. This radiation reaches the outer shell of the space station and is absorbed by it. Therefore, the total heat provided to the workstation per unit time is the sum of the heat Q released during the operation of the device and the heat Q absorbed by the inner surface of the screen, that is, H. Equal to $2Q$.According to the heat balance conditions, the same amount of heat must be released, so
$\dfrac{Q}{{2Q}} = \dfrac{{{T^4}}}{{T_X^4}}$
where ${T_x}$ is the required temperature of the envelope of the station.
As, $T = 500K$, then ${T_x} = {2^{\dfrac{1}{4}}}T$
${T_x} = {2^{\dfrac{1}{4}}}(500) \\\ \therefore {T_x} = 594.6$

Hence, the ${T_x}$ will be $594.6\,K$.

Note: As, $q = \alpha {R^3}$and since the heat released per unit area of ​​this area is proportional to ${T^4}$ , and all the heat released in the equilibrium state is released into the room, we can write $q = \beta {R^2}{T^4}$ (the area of is the same as ${R^2}$ is proportional, and $\beta$ is the coefficient) by equating these two expressions to q, we get ${T^4} = \dfrac{\alpha }{\beta }R$.