If the engine takes in an amount of heat Q at temperature Tin, the entropy of the environment will decrease by:
-Q/Tin
If the engine performs an amount of work W, then in needs to dump an amount of heat of Qout = Q -W to make the energy balance neutral. If it does this at a temperature of Tout, that will contribute to an entropy rise of the environment of:
Qout/Tout = (Q-W)/Tout
The total entropy change of the environment is thus:
Delta S = -Q/Tin + (Q-W)/Tout
If the engine's internal state is the same after each cycle, then Delta S is the total entropy change and this has to be equal or larger than zero. So, we have:
-Q/Tin + (Q-W)/Tout >= 0
W/Q is the efficiency eta, dividing by Q gives:
eta <= 1 - Tout/Tin ------->
Tout/Tin <= 1 - eta ------->
Tin >= Tout/(1 - eta)
The maximum possible efficiency of a heat engine which exhaust its heat at a temperature of 46.0o is 39.0 percent. What is the minimum value of the temperature at which the engine takes in heat?
*I think I keep doing a step wrong in here so step by step instruction would be so helpful!
Thanks in advance :)
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