AIPMT | 2014CalorimetryMedium
Question
Steam at 100oC is passed into 20 g of water at 10oC. When water acquires a temperature of 80oC, the mass of water present will be:
[Take specific heat of water = 1 cal g-1 oC-1 and latent heat of steam = 540 cal g-1]
[Take specific heat of water = 1 cal g-1 oC-1 and latent heat of steam = 540 cal g-1]
Options
A.24 g
B.31.5 g
C.42.5 g
D.22.5 g
Solution
Heat gain by water = Heat lost by steam
20 × 1 × (80 - 10) = m × 540 + m × 1 × (100 - 80)
⇒ 1400 = 560 m
⇒ m = 2.5 g
Total mass of water = 20 + 2.5 = 22.5 g
20 × 1 × (80 - 10) = m × 540 + m × 1 × (100 - 80)
⇒ 1400 = 560 m
⇒ m = 2.5 g
Total mass of water = 20 + 2.5 = 22.5 g
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