An icebox almost completely filled with ice at $˚ 0 C$ is dipped into a large volume of water at $2 ˚ 0 C$ . The box has walls of surface area $2400 {c m}^{2}$ , thickness $2.0 m m$ and thermal conductivity $0.06 W / m - C$ . Calculate the rate at which the ice melts in the box. Latent heat of fusion of ice = $3.4 \times 10^{5} J / k g$

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Solution

$A = 2400 c m^{2} = 2400 \times 10^{- 4} m^{2}$
$t = 2 m m = 2 \times 10^{- 3} m$
$K = 0.06 w / m -^{0} C$
$θ_{1} = 20^{0} C$
$θ_{2} = 0^{0} C$
$\frac{Q}{t} = \frac{K A (θ_{1} - θ_{2})}{l} = \frac{0.06 \times 2400 \times 10^{- 4} \times 20}{2 \times 10^{- 3}} = 24 \times 6 \times 10^{- 1} \times 10 = 24 \times 6 = 144 J / s e c$
Rate in which ice melts = $\frac{m}{t} = \frac{Q}{t \times L} = \frac{144}{3.4 \times 10^{5}} k g / h = \frac{144 \times 3600}{3.4 \times 1065} K g / s = 1.52 K g / s .$

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An icebox almost completely filled with ice at ˚0C is dipped into a large volume of water at 2˚0C. The box has walls of surface area 2400cm2, thickness 2.0mm and thermal conductivity 0.06W/m−C. Calculate the rate at which the ice melts in the box. Latent heat of fusion of ice = 3.4×105J/kg

A=2400cm2=2400×10−4m2t=2mm=2×10−3mK=0.06w/m−0Cθ1=200Cθ2=00CQt=KA(θ1−θ2)l=0.06×2400×10−4×202×10−3=24×6×10−1×10=24×6=144J/secRate in which ice melts = mt=Qt×L=1443.4×105kg/h=144×36003.4×1065Kg/s=1.52Kg/s.