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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