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10-1.Thermometry, Thermal Expansion and Calorimetry
hard
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 $90°C$ ........ $gm$ (Latent heat of steam is $540\, cal/gm)$ is
A
$24.8$
B
$24$
C
$36.6$
D
$30$
Solution

(a) 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
==> $m \times 540 + m \times 1 \times (100 – 90) = 22 \times 1 \times (90 – 20)$
==> $m = 2.8$ $gm$
The net mass of the water present in the mixture $ = 22 + 2.8 = 24.8\,gm.$
Standard 11
Physics
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