IIT JEE Advanced 2006 Paper 1

The heat lost by steam when it condenses into water (from 100°C to 0°C) is
Q1 = mLv + msΔT
= [(0.05 × 540) + (0.05 × 1 × 100)] × 103
= 32 kcal
The heat required by ice to reach 0°C and then melt (at 0°C) is
Q2 = m′s′ΔT ′ + m′Lf
= [(0.45 × 0.5 × 20°) + (0.45 × 80)] × 103
= 40.5 kcal
That is, Q2 > Q1. Hence entire amount of ice cannot melt; hence,
the temperature of the mixture remains at 0°C.