If 4 , kg of ice is inside a closed cubical t | vedclass

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(A) The rate of heat flow $dQ/dt$ through the walls of the box is given by $dQ/dt = (K \cdot A \cdot \Delta T) / d$. Here, the surface area $A = 6 	i

Vedclass · Accepted Answer

$3.678$

Vedclass · Answer

(A) The rate of heat flow $dQ/dt$ through the walls of the box is given by $dQ/dt = (K \cdot A \cdot \Delta T) / d$. Here, the surface area $A = 6 	imes (side)^2 = 6 	imes (0.2 \, m)^2 = 6 	imes 0.04 = 0.24 \, m^2$. The thickness $d = 0.04 \, m$ and temperature difference $\Delta T = 50^{\circ}C - 0^{\circ}C = 50^{\circ}C$. So, $dQ/dt = (0.01 	imes 0.24 	imes 50) / 0.04 = 0.12 / 0.04 = 3 \, Js^{-1}$. Total time $t = 10 \, hours = 10 	imes 3600 = 36000 \, s$. Total heat gained $Q = (dQ/dt) 	imes t = 3 	imes 36000 = 108000 \, J$. Mass of ice melted $m_{melted} = Q / L = 108000 / (335 	imes 10^3) \approx 0.322 \, kg$. Mass of ice remaining $= 4 \, kg - 0.322 \, kg = 3.678 \, kg$.

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