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Question

An electric heater is used in a room of total wall area 137 m2 to maintain a temperature of 20C inside it when the outside temperature is 10C. The walls have three layers of different materials. 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 cm. Find the power of the electic heater. Assume that there is no heat loss through the floor and ceiling. The thermal conductivities of wood, cement and brick are 0.125 W/mC, 1.5 W/mC and 1.0 W/mC respectively.

A
9 W
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B
22.5 W
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C
9 kW
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D
22.5 kW
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Solution

The correct option is C 9 kW
Thermal resistance offered by the walls in series combination is given by
R=R1+R2+R3
i.e L1+L2+L3KeqA=L1K1A+L2K2A+L3K3A


From the diagram, we can write the above equation as,
L1+L2+L3Keq=L1K1+L2K2+L3K3
From the data given in the question, (2.5+1+25)×102Keq=(2.50.125+11.5+251)×102
Keq=0.624 W/mC
The power of the heater H=KeqAΔTΔx
=0.624×137(3028.5×102)9 kW
Thus, option (c) is the correct answer.

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