First slide

First law of thermodynamics

Question

If 1 ${cm}^{3}$ of water is vaporized (latent heat of vaporization = 540 $cal / g^{0} C$ ) at P = 1 atm. If the volume of steam formed is 1671 ${cm}^{3}$ , calculate increase in internal energy.

Moderate

A

204 cal
B

167 cal
C

540 cal
D

373 cal

Solution

$∆ Q = ∆ U + ∆ W$
m = 1 gm
$L_{v} = 540 cal / gm$
$∆ Q = 1 \times 540 = 540$
$540 = ∆ U + P ∆ V$
$540 = ∆ U + 10^{5} \times (1671 - 1) \times 10^{- 6}$
540 = $∆ U + 167$
$∆ U = 540 - 167 = 373 cal$

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