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(D) The system consists of three layers in series: glass, air, and glass. The equivalent thermal resistance $R_{eq}$ is given by $R_{eq} = R_1 + R_2 +

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(D) The system consists of three layers in series: glass, air, and glass. The equivalent thermal resistance $R_{eq}$ is given by $R_{eq} = R_1 + R_2 + R_3$, where $R = \frac{l}{kA}$.Given: $l_1 = l_3 = 1 \,cm = 0.01 \,m$, $l_2 = 5 \,cm = 0.05 \,m$, $A = 2.6 \,m^2$, $k_{glass} = 0.8 \,Wm^{-1} K^{-1}$, $k_{air} = 0.08 \,Wm^{-1} K^{-1}$.$R_1 = R_3 = \frac{0.01}{0.8 	imes 2.6} = \frac{0.01}{2.08} \approx 0.0048 \,K/W$.$R_2 = \frac{0.05}{0.08 	imes 2.6} = \frac{0.05}{0.208} \approx 0.2404 \,K/W$.$R_{eq} = 2 	imes \left(\frac{0.01}{2.08}ight) + \frac{0.05}{0.208} = \frac{0.02}{2.08} + \frac{0.5}{2.08} = \frac{0.52}{2.08} = 0.25 \,K/W$.The rate of heat flow $H = \frac{\Delta T}{R_{eq}} = \frac{18 - (-2)}{0.25} = \frac{20}{0.25} = 80 \,W$.

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