$(a)\;40^{0}C\qquad(b)\;60^{0}C\qquad(c)\;80^{0}C\qquad(d)\;50^{0}C$

Answer : $\;50^{0}C$

Explanation :

Rate of heat flow is the same

$\large\frac{K_{1}A(T_{1}-\theta)}{L_{1}} = \large\frac{K_{2}A (\theta - T_{2})}{L_{2}}$

$\large\frac{0.9 A (100 - theta)}{100} = \large\frac{0.18 A (\theta - 0^{0})}{20}$

$90 - 0.9 \theta = 0.9 \theta$

$1.8 \theta = 90$

$\theta = \large\frac{90}{1.8} = 50^{0}C$

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