Question

A lump of 0.10 kg of ice at -10°C is put in 0.15 kg of water at 20°C.How much water and ice will be found in the mixture when it has reached thermal equilibrium? Specific heat capacity of ice=0.50 kcal/kg and its latent heat of melting=80 kcal/kg

Answer 1

Heat taken by 0.10 kg of ice in coming from – 10ºC to 0ºC
= 100 × 0.5 × 10 = 500 cal.
If m gm of ice melts, then heat taken in melting = m × 80 cal.
Heat given by water = 150 × 20 = 300 cal.
∴ 500 + m × 80 = 3000 or m = (2500/80) = 31.25 gm
∴ Ice present at thermal equilibrium = 100 – 31.25 = 67.75 gm.
Water present at thermal equilibrium = 150 + 31.25 = 181.25 gm