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Radhika Batra Grade: 11
        An 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-°C. Calculate the rate at which the ice melts in the box. Latent heat of fusion of ice = 3.4 × 105 J/kg.
3 years ago

Answers : (1)

Kevin Nash
askIITians Faculty
332 Points
										Sol. A = 2400 cm2 = 2400 × 10^–4 m^2

ℓ = 2 mm = 2 × 10^–3 m
K = 0.06 w/m-°C
θ base 1 = 20°C
θ base 2 = 0°C
Q/t = KA(θ base 1 – base 2)/ℓ = 0.06 * 2400 * 10^-4 * 20/2 * 10^-3 = 24 * 6 * 10^-1 * 10 = 24 * 6 = 144 j/sec
Rate in which ice melts = m/t = Q/t * L = 144/3.4 * 10^5 Kg/h = 144 * 366/3.4 * 10^4 Kg/s = 1.52 kg/s.
3 years ago
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