Heat capacitance

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Homework Statement



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.


Homework 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


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.)
 

Answers and Replies

  • #2
[tex]W_{b}= \int Pdv[/tex]

where pressure is constant and [tex]dv[/tex] is the change in volume. The volume goes from .01m^3 to .824m^3.

This will give you the work done BY the system if it was in a piston cylinder device.

To computer the work required to vaporize 1kg of water, simply multiply the the heat of vaporization by the amount of water.

[tex]2.2e6 kJ/kg * 1kg = 2200 kJ[/tex]

Using your own formula for the first law of thermo:

[tex]\Delta h = 2200 kJ - 164.6 kJ = 2035 kJ[/tex]
 

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