What Happens to dU, dQ, and dW During Boiling?

[prasannaworld]

Boiling Internal Energy - URGENT

A quick question:

dU = dQ - dW

What are dU, dQ and dW during Boiling?

I thought:
-the particles do work, hence dW is +
-dU increases as particles gain more energy

so dQ = dU + dW - showing that heat flows in. Which made sense to me.

Or is Boiling Isothermal?

 

[vanesch]

It is difficult to say, because you didn't specify the way things boil. First of all, you shouldn't look at *microscopic* work ("the particles do work"). The whole idea of thermodynamics is that we can take abstraction from any microscopic model. So if you boil water in an open kettle, then yes, the steam does work: it pushes away the air that was there, at 1 atmosphere. You can picture this by considering the boiling water in a cylinder with a piston. The steam will push the piston out, doing work against the pressure of the air.

However, there's another change too. The internal energy of steam is larger than the internal energy of liquid water at the same temperature. So dU is positive (liquid water (lower U) changes into steam (higher U)). dW is positive (the system DOES work on the air), and of course, dQ is positive (heat is given to the system).

If the water is in a closed volume, then the steam doesn't do any work. So now, dW is 0. But dU is still positive (the change in internal energy between water and steam). And dQ remains positive (we're heating the water).

The difference between the U of water and the U of steam is called the "latent heat".