Q. 65.0( 2 Votes )

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–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.

Answer :


Given:
Temperature of the water : T1 = 20° C
Temperature of the ice box : T2= 0° C
Surface area of the box : A = 2400 cm2 = 0.24 m2
Thickness of the box : x = 2 mm = 0.002 m
Thermal conductivity of the box : K = 0.06 W m–1 °C–1.
Latent heat of fusion of ice = 3.4 × 105 J kg–1.
Formula used:
Rate of amount of heat flowing is given as:

Here, Δθ is the amount of heat transferred, ΔT is the temperature difference, K is the thermal conductivity of the material, A is the area of cross section of the material and x is the thickness of the material.
Also,
Δθ = Q =L× m
Here, Q is the amount of heat absorbed or released, L is the Latent heat and m is the mass of the substance.
We need to find rate at which ice melts, which means we need to calculate the decrease in mass of the ice per second: Δm/Δt
Where Δm is the rate of change of mass.
Now,


Substituting for Δθ we get



OR

Hence, the rate at which ice melts in the box is
4.23× 10-4 kg/s or 1.52 kg/hr.


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