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Question
A icebox almost completely filled with ice at 0°C is dipped into a large volume of water at 20°C. The box has walls of surface area 2400 cm2, thickness 2.0 mm and thermal conductivity 0.06 W m−1°C−1. Calculate the rate at which the ice melts in the box. Latent heat of fusion of ice = 3.4 × 105 J kg−1.
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Solution
Area of the walls of the box, A = 2400 cm2 = 2400 × 10−4 m2
Thickness of the ice box, l = 2 mm = 2 × 10−3 m
Thermal conductivity of the material of the box, K = 0.06 W m−1 °C−1
Temperature of the water outside the box, T1 = 20°C
Temperature of ice, T2 = 0°C
`\text{ Rate of flow of heat }` = `\text{ Temprature differences }/ \text{ Thermal resistance}`
`⇒ (ΔQ)/(Δt) = ( T_1 - T_2 ) / ( l/(KA)`
`⇒ (DeltaQ)/(Deltat) = (20/ 2 xx 10 ^-3) xx 0.06 xx 2400 xx 10^-4`
`⇒ (DeltaQ)/(Deltat) = 24 xx 6 = 144` J/s
Rate at which the ice melts = `(mL_f)/t`
`⇒ (DeltaQ)/(Deltat) = (m/t) L_f`
`⇒ 144 = (m/t) xx 3.4 xx 10^5`
`⇒ m/t = 144/ (3.4 xx 10^5 )` kg/s
`⇒ m/t = (144 xx 60 xx 60 )/( 3.4 xx 10^5)` Kg/s
`⇒m/t = 1.52` kg / h
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