Answer:
0.05 W/m°C
Explanation:
Surface area of box, A = 1.47 m²
thickness of wall, d = 3.91 cm = 0.0391 m
Power of heater, P = 15.9 W
difference in temperature, ΔT = 8.4 °C
Let K is the thermal conductivity.
Heat flows per unit time is given by
[tex]H=\frac{KA\Delta T}{d}[/tex]
[tex]15.9=\frac{K\times 1.47\times 8.4}{0.0391}[/tex]
K = 0.05 W/m°C
Thus, the thermal conductivity of the box is 0.05 W/m°C.
