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

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

$$\begin{gathered} \mathrm{Rate}\mathrm{of}\mathrm{flow}\mathrm{of}\mathrm{heat}=\frac{\mathrm{Temperature}\mathrm{difference}}{\mathrm{Thermal}\mathrm{resistance}} \\ \Rightarrow \frac{\Delta Q}{\mathrm{\Delta }t}=\frac{T_1-T_2}{{\displaystyle \frac{l}{kA}}} \\ \Rightarrow \frac{\Delta Q}{\mathrm{\Delta }t}=\left(\frac{20}{2\times {10}^{-3}}\right)\times 0.06\times 2400\times {10}^{-4} \\ \Rightarrow \frac{\Delta Q}{\mathrm{\Delta }t}=24\times 6=144\mathrm{J}/\mathrm{s} \end{gathered}$$
Rate at which the ice melts =$\frac{m{L}_f}{t}$
$$\begin{gathered} \Rightarrow \frac{\Delta Q}{\mathrm{\Delta }t}=\left(\frac{m}{t}\right)L_f \\ \Rightarrow 144=\left(\frac{m}{t}\right)\times 3.4\times {10}^5 \\ \Rightarrow \frac{m}{t}=\frac{144}{3.4\times {10}^5}\mathrm{kg}/\mathrm{s} \\ \Rightarrow \frac{m}{t}=\frac{144\times 60\times 60}{3.4\times {10}^5}\mathrm{kg}/\mathrm{h} \\ \Rightarrow \frac{m}{t}=1.52\mathrm{kg}/\mathrm{h} \end{gathered}$$