Question

46. Steam at 100°C is passed into 22 grams of
water at 20°C. When resultant temperature is
90°C, then weight of the water present is
1) 27.33 g
2) 24.8 g
3) 2.8 g
4) 30 g​

Answer 1

Answer:

The weight of the water present is 24.8 g

Explanation:

Let the mass of steam condensed be m.

The heat released by steam is sum of heat released while condensing form stem to ice and heat released while temperature is dropped from 100°C to 90°C.  

So,

Q₁  = (m×L) + (m×(100−90)×1)

⇒ Q₁ = 540 m + 10 m

⇒ Q₁ = 550 m

Heat absorbed by 22 g of water to increase its temperature from 20°C to 90°C is :

Q₂ = 22 × (90−20) × 1

⇒ Q₂ = 1540 cal

Equating the two heats,

m = 1540/550

⇒ m = 2.8 g

So,

Total mass = 22+2.8

⇒ Total Mass = 24.8 g