NEET-XII-Physics
28: Heat Transfer
- #12A cubical box of volume 216 cm3 is made up of 0.1 cm thick wood. The inside is heated electrically by a 100 W heater. It is found that the temperature difference between the inside and the outside surface is 5°C in steady state. Assuming that the entire electrical energy spent appears as heat, find the thermal conductivity of the material of the box.Ans :
Volume of the cube, V = a3 = 216 cm3
Edge of the cube, a = 6 cm
Surface area of the cube = 6a2
= 6 (6 × 10-2)2
= 216 × 10-4 m2
Thickness, l = 0.1 cm = 0.1 × 10-2 m
Temperature difference, Δ T = 5°C
The inner surface of the cube is heated by a 100 W heater.
∴ Power, P = 100 W
Power = Energy per unit time
∴ Rate of flow of heat inside the cube, R = 100 J/s
Rate of flow of heat is given by
`` \frac{\,\mathrm{\,\Delta Q\,}}{\,\mathrm{\,\Delta \,}t}=\frac{\Delta \,\mathrm{\,T\,}}{{\displaystyle \frac{l}{\,\mathrm{\,kA\,}}}}``
`` 100=\frac{\,\mathrm{\,k\,}\times 216\times {10}^{-4}\times 5}{0.1\times {10}^{-2}}``
`` \,\mathrm{\,k\,}=0.9259\,\mathrm{\,W\,}/\,\mathrm{\,m\,}°\,\mathrm{\,C\,}``
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