Question

A room at $20^{\circ }\text{C}$ is heated by a heater of resistance $\text{20 ohm}$ connected to $\text{200 V}$ mains. The temperature is uniform throughout the room and the heat is transmitted through a glass window of area $1{\text{m}}^2$ and thickness $\text{0}\text{.2 cm}$ . Calculate the temperature outside. Thermal conductivity of glass is $0.2\text{ cal}/{\text{m}}^{\circ }\text{C }\text{s}$ and mechanical equivalent of heat is $\text{4}\text{.2 J/cal}$ .

• A. $13.69^{\circ }C$
• B. $15.24^{\circ }C$
• C. $17.85^{\circ }C$
• D. $19.96^{\circ }C$

Answer 1

Answer: (B)

If $\theta$ is the temperature of outside, heat passing per second through the glass window,
$\frac{\mathrm{dQ}}{\mathrm{dt}}=\mathrm{KA}\frac{\left({\theta }_1-{\theta }_2\right)}{\mathrm{L}}$
$=\frac{\text{0.2}\times 1\times \left(20-\theta \right)\text{cal}}{\text{0.2}\times 10^{-2}}=100\left(20-\theta \right)$
And heat produced per second by the heater in the room
$P=\frac{V^2}{R}\frac{J}{S}=\frac{V^2}{R}\frac{cal}{s}$
$=\frac{200\times 200}{20\times \text{4.2}}=\text{476.2}\frac{\text{cal}}{\text{s}}$
Now as the temperature of the room is constant, the heat produced per second by heater must be equal to the heat conducted through the glass window.
$100(20-\theta )=476.2;\theta =15.24^{\circ }\mathrm{C}$