The discussion focuses on maximizing work output from a Carnot cycle operating between a finite reservoir (tank water) and an infinite reservoir (river water). The efficiency of the engine is contingent on the temperature of the tank water, which thermalizes to 10°C with the river. The correct equation for entropy change in this adiabatic system is $$\Delta S_{tank\ water}+\Delta S_{river\ water}=0$$, ensuring no entropy generation. Maximum work is achieved when the heat engine operates at maximum efficiency, calculated using $$dW=\eta~dQ=\eta~ m~c~dT$$.
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You left out the heat capacity of the water and you lost the minus sign.so_gr_lo said:
Yes, if ##T_{tank\ water}## is the initial value of the tank water temperature. How much heat does the tank water lose to the engine? What is the heat lost by the tank water to the engine minus the heat rejected by the engine to the river water?so_gr_lo said:
It's the same number I got through a different method that I outlined in post #2. Here is what I did. Maximum work is achieved when the heat engine has maximum efficiency which means operating a Carnot cycle between the variable tank temperature ##T## and the constant river temperature ##T_R##. When heat ##dQ## is transferred from tank to river, the work done by the Carnot engine is $$dW=\eta~dQ=\eta~ m~c~dT=m~c\left(1-\dfrac{T_R}{T}\right)dT.$$ Then, $$W=m~c\int_{T_0}^{T_R}\left(1-\dfrac{T_R}{T}\right)dT=m~c\left[ (T_0-T_R)+T_R\ln\left(\frac{T_R}{T_0}\right) \right].$$so_gr_lo said: