See attached file. Thank you so much! Bye© BrainMass Inc. brainmass.com March 4, 2021, 8:41 pm ad1c9bdddf
Suppose we have an engine that converts heat into work. We assume that the engine's thermodynamic state goes through cycles so that it returns to the same state periodically. The engine's work output and heat intake can then be defined by looking at the energy balance over the course of one cycle. If an engine has some hidden stored energy like gas under pressure, then it could do a lot of work by letting that gas expand while not taking in much heat. But that would be cheating. Since the engine's thermodynamic state would not be the same after the gas has expanded as before, we would disqualify that result. The criterium that the engine must return to the same state ensures that the engine is operated in a way that is sustainable on the long run.
If an engine absorbs an amount Q of heat at some temperature T1, then the entropy of the engine will increase by at least Q/T. Unless at some point some heat is dumped, the entropy of the engine cannot return to what is was. So, the work done by the engine would not qualify. This means that the engine must dump some heat Q' at temperature T2 so that the total entropy change of the engine equals zero. In the reversible case the entropy change of the engine is given by:
Delta S = Q/T1 - Q'/T2
Since Delta S must be zero, we have:
Q' = T2/T1 Q
The internal energy of the engine must be the same as what is was before, this means that the engine will perform an amount of work equal to W = Q - Q' = Q (1 - T2/T1) = Q (T1 - T2)/T1.
In general we have
Delta S >= Q/T1 - Q'/T2
If we then ...
A detailed solution is given