(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

When water is boilded under a pressure of 2 atm, the heat of vaporization is 2.2 x 10^6 J/kg and the boiling point is 120 degree C. At this pressure, 1 kg of water has a volume of 10^-3 cubic meter and 1 kg of steam has a volume of 0.824 cubic meter.

a.) Compute the work done when 1 kg of steam is formed at this temperature.

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

2. Relevant equations

boiling point = 393 K

number of moles = 1000 / 18.0

first law of thermo: change in internal energy = energy input - work done by system

3. The attempt at a solution

a.) W = p(V2-V1) since it is isobaric

= 2x10^5 (0.824 - 10^-3)

= 1.646 x 10^5 J (work done by water)

(Is this correct?)

b.) pV = nRT

2x10^5x10^-3 = (1000/18.0) x 8.31 x T

T = 0.43 K

therefore: change in T = 393 - 0.43

= 392.57 K

(Don't I need to know the heat capacitance to work out the heat used to increase the temperature of the water to the boiling point? If I find this, then I can simply add this to the heat of vaporization to get the total heat input. Then, from first law of thermodynamics, I can find the change in internal energy.)

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Heat capacitance

**Physics Forums | Science Articles, Homework Help, Discussion**