# Carnot Engine Theory Question

1. Jan 19, 2010

### tiredryan

Note this is more of a coursework question about the theory behind Carnot engines as opposed to an actual homework question.

1. The problem statement, all variables and given/known data
3. The attempt at a solution

So I am reading the theory behind a Carnot Engine and I am a little confused about the cooling stages of the Carnot cycle. This is steps 3 and 4 called, "Reversible isothermal compression of the gas at the "cold" temperature, TC" and "Isentropic compression of the gas(isentropic work input)" at http://en.wikipedia.org/wiki/Carnot_cycle#The_Carnot_cycle.

From my reading, it seems to be that some work has to be done on the system to return the piston to its lower volume stage. Basically part of the work that was created to push the piston out is used to return the piston back in. Is this a correct understanding?

If so, I am confused when thinking about the following thought experiment. Lets imagine that I had a balloon filled with gas connected to a shaft connected to an object I wanted to apply work on. When I place the balloon in a hot bath the balloon expands and the shaft moves and the object moves. The the shaft pushes the object over a distance and work is done on the object. When I switch out the balloon into cold water, the balloon shrinks and the object moves back. The object is moved by the force applied from the shaft and work is done on the object again. So in the forward cycle and the backward cycle work is being done to push the object. When I move the balloon cold water, I do not need to apply any addition work, but rather work is being done to the object of interest.

I am not sure where my confusion stems. Thanks in advance.

Last edited: Jan 19, 2010
2. Jan 19, 2010

### vela

Staff Emeritus
Essentially yes. Work has to be done to restore the system back to its initial state, but it doesn't necessarily have to be part of the work output from the earlier part of the process.

It's implied in your description that the object comes to rest in between the two phases. As it decelerates, it's doing work on the gas. Also, when the gas cool, it will contract only if something outside does work. In this case, both the balloon and the outside atmosphere will perform work, and part of the work goes into accelerating the object.