91Ó°ÊÓ

A cubical box side$=20cm$

Thickness$=1.2cm$

Power of lightbulb $=100W$

Temperature$={20}^{∘}C$

What we have is a thin cube, the inner side of which is at a temperature $T_i$, the temperature we are looking for, and the outer side is at the same temperature of the air, $T_a$. The power transmitted through the concrete is $P$, the power of the lamp, since the temperature inside is constant; that is, the air inside is not getting warmer nor colder. Therefore, we have

$P=\frac{kA}{L}\left(T_i-T_a\right)$

We can express the temperature inside as

$T_i-T_a=\frac{PL}{kA}⇒T_i=\frac{PL}{kA}+T_a$

We know that the thickness of the concrete is $L=0.012\mathrm{m}$and the thermal conductivity constant for concrete, referring table $19.5$on the book, is $0.8\frac{\mathrm{W}}{\mathrm{mK}}$.

We can now substitute, knowing that the area of a cube is $A=6a^2$where $a$is the side length, to find

$T_1=\frac{100·0.012}{0.8·6·0.2^2}+20=26.25\mathrm{C}$

Since the quantity $\frac{\text{PL}}{kA}$ expresses temperature difference and not absolute temperature, so substituting in degrees Celsius,.

Therefore, the air temperature inside the box when the light is$26.25C$.