A room is maintained at $20^{\circ} \mathrm{C}$ by a heater of resistance 20 ohm connected to 200 vo

Question

A room is maintained at $20^{\circ} \mathrm{C}$ by a heater of resistance 20 ohm connected to 200 volt mains. The temperature is uniform throughout the room and the heat is transmitted through a glass window of area $1 \mathrm{~m}^{2}$ and thickness 0.2 cm . The temperature outside in $^{\circ} \mathrm{C}$ (Nearest Integer) . Thermal conductivity of glass is $0.2 \mathrm{cal} \mathrm{m}^{-1} \mathrm{~s}^{-1}\left({ }^{\circ} \mathrm{C}ight)^{-1}$ and mechanical equivalent of heat is $4.2 \mathrm{~J} \mathrm{cal}^{-1}$.\

Professor PC Thomas Classes · Answer

Heat produced by the heater per second is\ $\frac{Q}{t}=\frac{V^{2}}{R}=\frac{(200)^{2}}{20}=2000 \mathrm{Js}^{-1}$ $$\begin{equation*} =\frac{2000}{4.2} \mathrm{cal} \mathrm{s}^{-1} 	ag{i} \end{equation*}$$ Heat transmitted through the window per second is $$ \frac{Q}{t}=\frac{K A\left(T_{1}-T_{2}ight)}{d} $$ where, $T_{2}=$ temperature outside.\ Given $T_{1}=20^{\circ} \mathrm{C}, d=0.2 \mathrm{~cm}$,\ $A=1 \mathrm{~m}^{2}=10^{4} \mathrm{~cm}^{2}$ and\ $K=0.2 \mathrm{cal} \mathrm{m}^{-1} \mathrm{~s}^{-1}\left({ }^{\circ} \mathrm{C}ight)^{-1}$\ $=0.002 \mathrm{cal} \mathrm{cm}^{-1} \mathrm{~s}^{-1}\left({ }^{\circ} \mathrm{C}ight)^{-1}$. Thus $$\begin{align*} \frac{Q}{t} & =\frac{0.002 	imes 10^{4} 	imes\left(20-T_{2}ight)}{0.2} \ & =100\left(20-T_{2}ight) \mathrm{cal} \mathrm{~s}^{-1} 	ag{ii} \end{align*}$$ Equating (i) and (ii), we have $$ 100\left(20-T_{2}ight)=\frac{2000}{4.2} $$ which gives $T_{2}=15.24^{\circ} \mathrm{C}$.\