Question

What mass of steam at 100∘C must be added to 1.10 kg of ice at 0∘C to yield liquid water at 19 ∘C? The heat of fusion for water is 333 kJ/kg , the specific heat is 4186 J/kg⋅C∘ , the heat of vaporization is 2260 kJ/kg .

Answer 1

Heat required for converting water to ice at 0°C,

= 1.10*333000 J

= 366300 J

Heat required for raising temperature of water from 0°C to 19°C,

= 1.10*4186*19

= 87487.4 J

Total heat required = 366300 + 87487.4 = 453787.4 J

Heat to be removed from steam at 100°C to convert into water at 100°C,

= 2260000*M

Heat to be removed to cool water at 100°C to water at 19°C

= M*4186*(100-19)

= 339066M

Total heat lost = (2260000 + 339066)*M = 2599066M

Total heat lost = total heat required

2599066*M = 453787.4

M = 453787.4/2599066

= 0.17 kg of steam