question_answer

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 materials. The innermost layer 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 \[1.0\operatorname{W}/m-{}^\circ C\] respectively.

A) \[900 W\]                      

B) \[9000 W\]

C) \[90 W\]            

D) \[1900 W\]

Correct Answer: B

Solution :

Equivalent thermal conductivity of the wall

\[K=\frac{{{\ell }_{1}}+{{\ell }_{2}}+{{\ell }_{3}}}{\frac{{{\ell }_{1}}}{{{K}_{1}}}+\frac{{{\ell }_{2}}}{{{K}_{2}}}+\frac{{{\ell }_{3}}}{{{K}_{3}}}}\]\[=\frac{0.025+0.01+0.25}{\left( \frac{0.025}{0.457}+\frac{0.01}{1.5}+\frac{0.025}{1.0} \right)}\]\[=\frac{0.285}{0.457}=0.624W/m-{}^\circ C\]

The rate of flow of heat is given by \[H=KA\frac{{{T}_{1}}-{{T}_{2}}}{L}\]

\[=0.624\times 137\times \frac{\left[ 20-\left( -10 \right) \right]}{0.285}\]       \[=\frac{0.624\times 137\times 30}{0.285}=9000\operatorname{W}.\]