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 \[H=KA\frac{{{T}_{1}}-{{T}_{2}}}{L}\] |
\[=0.624\times 137\times \frac{\left[ 20-\left( -10 \right) \right]}{0.285}\] \[=\frac{0.624\times 137\times 30}{0.285}=9000\operatorname{W}.\] |
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