Steam is passed into 22 g of water at 20 °C. the mass of water that will be present when the water acquires a temperature of 90 °C

Q: Steam is passed into 22 g of water at 20 °C. the mass of water that will be present when the water acquires a temperature of 90 °C (Latent heat of steam is 540 cal/g) is

(a) 24.8 g

(b) 24 g

(c) 36.6 g

(d) 30 g

Ans: (a)

Sol: m L_s + m s_w (100-90) = 22 s_w (90-20)

m × 540 + m × 1 × 10 = 22 × 70

m = (22 × 70)/550 = 2.8 g

total mass of water = 22 + 2.8 = 24.8 g