Question

When water is boiled under a pressure of 2.00atm, the heat of vaporization is 2.20×106J/kg and the boiling point is 120∘C. At this pressure, 1.00kg of water has a volume of 1.00×10−3m3, and 1.00 kg of steam has a volume of 0.824m3.
a)Compute the work done when 1.00kg of steam is formed at this temperature

b)Compute the increase in internal energy of the water.

Answer 1

Answer:

2033219.05 J

Explanation:

V = Volume

P = Pressure = 2 atm

m = Mass of water = 1 kg

$L_v$ = Heat of vaporization = $2.2\times 10^6\ J/kg$

Work done in an isobaric system is given by

$W=-P\Delta V\\\Rightarrow W=-2\times 101325\times (1\times 10^{-3}-0.824)\\\Rightarrow W=166780.95\ J$

Work done is 166780.95 J

Change in internal energy is given by

$\Delta U=Q-W$

Heat is given by

$Q=mL_v\\\Rightarrow Q=1\times 2.2\times 10^6\\\Rightarrow Q=2.2\times 10^6\ J$

$\Delta U=2.2\times 10^6-166780.95\\\Rightarrow \Delta U=2033219.05\ J$

The increase in internal energy of the water is 2033219.05 J