Properties Of Solids And Liquids Question 120

Question: Steam is passed into 22 gm 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/gm) is [SCRA 1994]

Options:

A) 24.8 gm

B) 24 gm

C) 36.6 gm

D) 30 gm

Show Answer

Answer:

Correct Answer: A

Solution:

Let m gm of steam get condensed into water (By heat loss).

This happens in following two steps.

Heat gained by water (20°C) to raise it’s temperature upto 90° =22×1×(9020)

Hence, in equilibrium heat lost = Heat gain

therefore m×540+m×1×(10090)=22×1×(9020)

therefore m=2.8 gm

The net mass of the water present in the mixture =22+2.8=24.8gm.



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