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 [Take specific heat of water = 1 cal g-1 C-1 and latent heat of steam = 540 cal g-1]
- 24 g
- 31.5 g
- 42.5 g
- 22.5 g
Answer
(4) 22.5 g
Solution
Let m g of steam condense. Heat lost by steam = heat gained by water
m 540 + m 1 (100 - 80) = 20 1 (80 - 10)
560 m = 1400
m = 2.5 g
Mass of water = 20 + 2.5 = 22.5 g
Answer: (4)