First slide

First law of thermodynamics

Question

A sample of 0.1 g of water at l00°C and normal pressure.(1.013 x 10⁵ Nm^-2 )requires 54 cal of heat energy to convert to steam at 100°C. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample, is:

Moderate

A

104.3 J
B

208.7 J
C

42.2 J
D

84.5 J

Solution

$Q = 54 cal = 54 \times 4.18 joule = 225.72 joule W = P [V_{steam} - V_{water}] [For water 0.1 gram = 0.1 cc] = 0.013 \times 10^{5} [167.1 \times 10^{- 6} - 0.1 \times 10^{- 6}] joule = 1.013 \times 167 \times 10^{- 1} = 16.917 joule By FLOT \Rightarrow ∆ U = Q - W = 225.72 - 16.917 = 208.8 joule .$

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