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

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?
 Recognitions: Science Advisor 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.

Recognitions:
Homework Help
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 ($\Delta S = 0$).