Question

Ten ice cubes of 2cm edge, at a temperature of -5°C are added to 1 liter of water at 20°C in a well-insulated thermos flask. What is the equilibrium temperature ? (Use: cp ice = 2.1 kJ/(kg °C), heat of fusion for ice: hice-water = 334kJ/kg, and cp water = 4.18 kJ/(kg °C)).

Answer 1

1 ice cube volume= 8 cm³ so, 10 ice cubes are 80 cm³.

Density of ice = 0.9 g/cm³ then mass of ice = 80 cm³ * 0.9 g/cm³ = 72g or 0.072 kg.

Heat absorbed by ice before melting = 0.072 kg* 2.1 kJ/kg °C * 5 ° C = 0.76 kJ

Heat absorbed during melting = 0.072 kg* 334 kJ/kg = 24.0 kJ

Let equilibrium temperature be x ° C.

Heat absorbed = Heat lost

24 kJ + 0.76 kJ + 0.072 kg* 4.18 kJ/kg °C * (x-0 )°C = 1 kg* 4.18 kJ/kg ° C * (20-x) °C

=> 24.76 kJ + 0.3x kJ = 83.6 kJ - 4.18x kJ

=> 4.48x = 58.84

=> 13.13 °C

The equilibrium temperature is 13.13°C.