Consider a sealed electronic box whose base dimensions are 40 cm by 40 cm and whose height is 20 cm The electronic box is placed in a large vacuum chamber such that there is no air around the box and

From the given information, we can use the Stefan-Boltzmann law to determine the rate of radiation heat transfer from the box:

$q'' = \epsilon \sigma (T_{sur}^4 - T_{box}^4)$

where $q''$ is the heat transfer per unit area, $\epsilon$ is the emissivity of the surface, $\sigma$ is the Stefan-Boltzmann constant, $T_{sur}$ is the temperature of the surrounding surfaces, and $T_{box}$ is the temperature of the outer surface of the box.

To determine the temperature of the surrounding surfaces, we need to assume a value for the temperature of the outer surface of the box. Let's start with an initial guess of $T_{box} = 40°C = 313.15K$, which is below the maximum allowed temperature of 60°C.

Using this value in the equation above, and solving for $T_{sur}$, we get:

$T_{sur} = \left( \frac{q''}{\epsilon \sigma} + T_{box}^4 \right)^{1/4}$

The heat transfer per unit area can be found by dividing the total power dissipated by the box by its surface area:

$q'' = \frac{100W}{(0.4m)^2} = 625 W/m^2$

Substituting all values into the equation, we get:

$T_{sur} = \left( \frac{625}{0.95 \cdot 5.67 \cdot 10^{-8}} + (313.15)^4 \right)^{1/4} = 308.7K$

Therefore, the temperature of the surrounding surfaces must be kept at 308.7K (35.6°C) or lower in order for the box to be cooled by radiation alone and maintain a maximum outer surface temperature of 60°C