Steam at 100°C is passed into 20 g of water at 10°C. When water acquires a temperature of 80°C, the mass of water present will be?

m(g)steam at 100°C → m(g)water at 100°C + 540m. (1)
m(g)water at 100°C → m(g)water at 80°C + (m)(1)(20). (2)
(1) + (2) m(g)steam at 100°C → m(g)water at 80°C + 560m(cal). (3)
20 g water at 10°C + (20)(1)70 → 20g water at 80°C. (4)
From (3) and (4) mix + 1400cal → (20+m)g water at 80°C + 560m(cal)
1400 = 560m
2.5 = m
Total mass of water present = (20 + 2.5)g = 22.5 g