# Entropy changes: Thermal energy to work and (some) back to Thermal energy

1. Oct 4, 2011

### kmarinas86

In classical thermodynamics, if we dissipated the kinetic energy of an object as thermal energy, then we would increase the entropy.

However, let's say we took 90% of some thermal energy in a reservoir, and converted it into work, and 10% of that is converted back into thermal energy after 1 minute is passed. This would mean that 81% of the thermal energy has been converted into work.

If we dissipate work as heat, entropy increases. So what happens if we convert heat into work? Shouldn't the opposite occur - a decline of entropy?

I think we should have a sum of changes, an entropy increase in excess of a subsequent decrease. Is this the correct view?

2. Oct 4, 2011

### atyy

In classical thermodynamics, you always need a temperature difference to do work. Every time you do work, you reduce the temperature difference. When a temperature difference is reduced, entropy increases.

3. Oct 4, 2011

### Andrew Mason

Just following on what atyy has said, you are correct that converting heat into work decreases the entropy of the surroundings (by -Qh/Th). The problem is that when you convert thermal energy into work in a heat engine, you have to expel a smaller amount of heat at a cooler temperature. The expelling of heat at a cooler temperature results in an increase in entropy of the surroundings of +Qc/Tc. The second law says that the net change in entropy Qc/Tc - Qh/Th can never be negative. A Carnot engine is the best you can do ($\Delta S = 0$).

So the second law puts a limit on the efficiency at which you can convert thermal energy into work. That upper limit is the Carnot engine cycle. It can never get better than that.

AM