Heat and Thermodynamics 3 Question 38
38. An electric heater is used in a room of total wall area $137 m^{2}$ to maintain a temperature of $+20^{\circ} C$ inside it, when the outside temperature is $-10^{\circ} C$. The walls have three different layers. The innermost layer is of wood of thickness $2.5 cm$, the middle layer is of cement of thickness $1.0 cm$ and the outermost layer is of brick of thickness $25.0 cm$. Find the power of the electric heater. Assume that there is no heat loss through the floor and the ceiling. The thermal conductivities of wood, cement and brick are $0.125,1.5$ and $1.0 W / m /{ }^{\circ} C$ respectively.
(1986, 8M)
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Answer:
Correct Answer: 38. $9091 W$
Solution:
- Let $R _1, R _2$ and $R _3$ be the thermal resistances of wood, cement and brick. All the resistances are in series. Hence,
$$ \begin{aligned} R & =R _1+R _2+R _3 \\ & =\frac{2.5 \times 10^{-2}}{0.125 \times 137}+\frac{1.0 \times 10^{-2}}{1.5 \times 137}+\frac{25 \times 10^{-2}}{1.0 \times 137}\left(\text { as } R=\frac{l}{K A}\right) \\ & =0.33 \times 10^{-2}{ }^{\circ} C / W \end{aligned} $$
$\therefore$ Rate of heat transfer,
$$ \begin{aligned} \frac{d Q}{d t} & =\frac{\text { temperature difference }}{\text { thermal resistance }}=\frac{30}{0.33 \times 10^{-2}} \\ & \approx 9091 W \end{aligned} $$
$\therefore$ Power of heater should be $9091 W$.
