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
Text Solution
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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° $= 22 \times 1 \times (90 - 20)$
Hence, in equilibrium heat lost = Heat gain
⇒ ⇒ $m \times 540 + m \times 1 \times (100 - 90) = 22 \times 1 \times (90 - 20)$
⇒ ⇒ $m=2.8^\circ$ gm
The net mass of the water present in the mixture = 22 + 2.8 = 24.8 gm.
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