Heat and Thermodynamics 3 Question 40
40. A room is maintained at $20^{\circ} C$ by a heater of resistance $20 \Omega$ connected to $200 V$ mains. The temperature is uniform throughout the room and the heat is transmitted through a glass window of area $1 m^{2}$ and thickness $0.2 cm$. Calculate the temperature outside. Thermal conductivity of glass is $0.2 cal m^{-1} s^{-1}\left({ }^{\circ} C\right)^{-1}$ and mechanical equivalent of heat is $4.2 Jcal^{-1}$.
(1978)
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Answer:
Correct Answer: 40. $15.24^{\circ} C$
Solution:
- Power produced by heater $=$ rate of heat flow through window
$$ \begin{aligned} & \therefore & \frac{V^{2}}{R} & =\frac{\text { Temperature difference }}{\text { Thermal resistance }}=\frac{20-\theta}{(l / K A)} \\ & \therefore & \theta & =20-\frac{V^{2} l}{K A R} \end{aligned} $$
Substituting the values we have
$$ \theta=20-\frac{(200)^{2}\left(0.2 \times 10^{-2}\right)}{(0.2 \times 4.2)(1)(20)}=15.24^{\circ} C $$
