Question

Steam at 100°C is passed into 20g 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 0} C^–1 and latent heat of steam = 540 cal g^–1 ]

(A) 24 g 

(B) 31.5 g 

(C) 42.5 g 

(D) 22.5 g

Answer 1

Correct option: (D) 22.5 g

Explanation:

m(g) steam at 100° → m(g) water at 100°C + 540m .....(A)

m(g) water at 100°C → m(g) water at 80°C + (m)(A) (20) .....(B)

(A) + (B)

m(g) steam at 100°C → m(g) water at 80° + 560m (cal) .....(C)

20 g water at 10°C + (20) (A) 70 → 20 g water at 80°C .....(D)

from (C) and (D)

mix + 1400 cal → (20 + m) g water at 80°C + 560m (cal)

1400 = 560m 

2.5 = m

Total mass of water present 

= (20 + 2.5)g 

= 22.5g