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(B) The rate of heat flow $dQ/dt$ through the walls is given by the formula: $dQ/dt = (K \cdot A \cdot \Delta T) / d$. Here, $K = 10^{-2} \,W m^{-1} K

Vedclass · Accepted Answer

$0.9$

Vedclass · Answer

(B) The rate of heat flow $dQ/dt$ through the walls is given by the formula: $dQ/dt = (K \cdot A \cdot \Delta T) / d$. Here, $K = 10^{-2} \,W m^{-1} K^{-1}$, $A = 1000 \,cm^2 = 0.1 \,m^2$, $\Delta T = 42^{\circ}C - 0^{\circ}C = 42 \,K$, and $d = 2 \,mm = 2 	imes 10^{-3} \,m$. Substituting these values: $dQ/dt = (10^{-2} 	imes 0.1 	imes 42) / (2 	imes 10^{-3}) = (4.2 	imes 10^{-2}) / (2 	imes 10^{-3}) = 21 \,W$ (or $21 \,J/s$). Total time $t = 160 \,minutes = 160 	imes 60 \,s = 9600 \,s$. Total heat transferred $Q = (dQ/dt) 	imes t = 21 	imes 9600 = 201600 \,J$. Mass of ice melted $m_{melted} = Q / L$, where $L = 336 	imes 10^3 \,J/kg$. $m_{melted} = 201600 / (336 	imes 10^3) = 0.6 \,kg$. Mass of ice remaining = Initial mass - Mass melted = $1.5 \,kg - 0.6 \,kg = 0.9 \,kg$.

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