Four 2 cm × 2 cm × 2 cm cubes of ice are taken out from a refrigerator and are put in 200 ml of a drink at 10°C. (a) Find the temperature of the drink when thermal equilibrium is attained in it. (b) If the ice cubes do not melt completely, find the amount melted. Assume that no heat is lost to the outside of the drink and that the container has negligible heat capacity. Density of ice = 900 kg m^{−3}, density of the drink = 1000 kg m^{−3}, specific heat capacity of the drink = 4200 J kg^{−1} K^{−1}, latent heat of fusion of ice = 3.4 × 10^{5} J kg^{−1}.

#### Solution

(a)

Given:-

Number of ice cubes = 4

Volume of each ice cube = (2 × 2 × 2) = 8 cm^{3}

Density of ice = 900 kg m^{−3}^{ }

Total mass of ice, m_{i} = (4 × 8 ×10^{−6} ×900) = 288×10^{−4} kg

Latent heat of fusion of ice, L_{i} = 3.4 × 10^{5} J kg^{−1}

Density of the drink = 1000 kg m^{−3}

Volume of the drink = 200 ml

Mass of the drink = (200×10^{−6})×1000 kg

Let us first check the heat released when temperature of 200 ml changes from 10^{o}C to 0^{o}C.

H_{w} = (200×10^{−6})×1000×4200×(10−0) = 8400 J

Heat required to change four 8 cm^{3} ice cubes into water (H_{i}) = m_{i}L_{i} = (288×10^{−4})×(3.4×10^{5}) = 9792 J

Since the heat required for melting the four cubes of the ice is greater than the heat released by water ( H_{i} > H_{w} ), some ice will remain solid and there will be equilibrium between

ice and water. Thus, the thermal equilibrium will be attained at 0^{o} C.

(b)

Equilibrium temperature of the cube and the drink = 0°C

Let M be the mass of melted ice.

Heat released when temperature of 200 ml changes from 10^{o}C to 0^{o}C is given by

H_{w} = (200×10^{−6})×1000×4200×(10−0) = 8400 J

Thus,

M×(3.4×10^{5}) = 8400 J

Therefore,

M = 0.0247 Kg = 25 g