How Does Boiling Water Affect Energy Calculations?

[edimeo25]

Homework Statement

A kettle containing 1kg of boiling water is heated until complete evaporation. Calculate q, w, "delta"U and "delta"H for this process. Suppose that the water vapor is an ideal gas and molar "delta"H of vaporisation at 373K = 40.6kJ/mol.

Homework Equations

H = U + PV
"delta"U = W + q

The Attempt at a Solution

number of moles = 1000g/(18.02g/mol) = 55.49mol

q = (molar "delta"H of vaporisation) x number of moles = 40.6x10^3J/mol x 55.49mol = 2.25x10^6J

U = H - PV
"delta"U = "delta"H - [("delta"P)(V) + ("delta"V)(P)] = "delta"H - [0 + nR"delta"T] = "delta"H
(since there is no pressure or temperature variation, "delta"U = "delta"H? I know pressure is constant, but is temperature constant too during this phase change?)

Therefore, "delta"U = 2.25x10^6J

W = "delta"U - q = 0J

I feel like I'm missing something in the calculation of "delta"U. Any help is greatly appreciated. Thanks.

 

[anathema]

Hello,

Thank you for your post. Your calculation for q and "delta"U are correct. However, there are a few things to note:

1. The temperature is not constant during the phase change from liquid to gas. In fact, it is the temperature at which the phase change occurs (in this case, 373K) that is used to calculate the molar "delta"H of vaporisation. So, in your calculation for q, you should use the temperature at which the phase change occurs, not the final temperature after complete evaporation.

2. Since there is no work being done during the phase change (the volume remains constant), the work done is indeed 0J. However, this does not mean that "delta"U = 0J. "Delta"U is the change in internal energy, and it takes into account both heat transfer and work done. In this case, since there is no work done, "delta"U is equal to q, which you have correctly calculated to be 2.25x10^6J.

3. Finally, to calculate the "delta"H, you can use the equation H = U + PV. Since the volume remains constant, PV = 0 and H = U. Therefore, "delta"H = "delta"U = 2.25x10^6J.

I hope this helps clarify the calculations. Let me know if you have any further questions. Good luck with your studies!

 

[baba89]

Your calculation for q and "delta"U are correct. However, for w, since the process is isothermal (constant temperature), the change in internal energy is equal to the work done by the system. Therefore, w = "delta"U = 2.25x10^6J. This means that all the energy input into the system is used to do work, and there is no heat transfer involved.

In terms of the temperature being constant during the phase change, it depends on the specific conditions of the system. In this case, since the water is being heated until complete evaporation, the temperature will remain constant at the boiling point (373K) until all the water has evaporated. However, if the system was not heated until complete evaporation, the temperature could vary during the phase change.