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.

Answer 1

The rate at which the ice melts in the box is $\frac{m}{t}=1.52 \mathrm{kg} / \mathrm{hr}$  

Explanation:

Step 1:

Given,  

Temperature of ice $\mathrm{T}_{1}=0^{\circ} \mathrm{C}$

Temperature of water $\mathrm{T}_{2}=20^{\circ} \mathrm{C}$

Surface area of wall of box $A=2400 \mathrm{cm}^{2}=0.24 \mathrm{m}^{2}$

Thickness of the wall of box $\mathrm{l}=2 \times 10^{-3} \mathrm{m}$

Thermal conductivity $\mathrm{K}=0.06 \mathrm{W} \mathrm{m}^{-1 \circ} \mathrm{C}^{-1}$

Latent heat of fusion of ice $\mathrm{L}=3.4 \times 105 \mathrm{J} \mathrm{kg}^{-1}$

Rate of heat flow   $\mathrm{q}=\frac{\Delta Q}{\Delta t}=\frac{\Delta T K A}{l}$

Step 2:

By substituting the values of given quantities in above equation we get

$q=144 \mathrm{J} / \mathrm{s}$

Step 3:

Also the rate of heat flow is given   $\frac{\Delta Q}{\Delta t}=\frac{m L}{t}$

By equating above equations we get  $\frac{m}{t}=144 / 3.4 \times 10^{5} \mathrm{kg} / \mathrm{s}$

$\frac{m}{t}=1.52 \mathrm{kg} / \mathrm{hr}$