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,
$\frac{\mathrm{dQ}}{\mathrm{dt}}=\mathrm{KA} \frac{\left(\theta_1-\theta_2\right)}{\mathrm{L}}$
$= \frac{\text{0.2} \times 1 \times \left(2 0 - \theta \right) \text{cal}}{\text{0.2} \times 1 0^{- 2}} = 1 0 0 \left(2 0 - \theta \right)$
And heat produced per second by the heater in the room
$P = \frac{V^{2}}{R} \frac{J}{S} = \frac{V^{2}}{R} \frac{c a l}{s}$
$= \frac{2 0 0 \times 2 0 0}{2 0 \times \text{4.2}} = \text{476.2} \frac{\text{cal}}{\text{s}}$
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.
$100(20-\theta)=476.2 ; \theta=15.24^{\circ} \mathrm{C}$