#### 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 cm^{2}, 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 × 10^{5} J kg^{−1}.

#### Solution

Area of the walls of the box, *A* = 2400 cm^{2} = 2400 × 10^{−4} m^{2}

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, *T*_{1} = 20°C

Temperature of ice, *T*_{2}_{ }= 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