The discussion revolves around maximizing work output in a heat engine, specifically analyzing a Carnot cycle operating between a tank of water and a river. Participants explore the relationship between temperature, efficiency, and work done by the engine as heat is transferred from the tank to the river.
The discussion is active, with participants questioning assumptions about temperature and entropy. Some guidance has been offered regarding the equations to use for entropy changes, and there is an ongoing exploration of how to calculate maximum work output based on varying temperatures.
Participants note the importance of heat capacity and the correct formulation of equations for entropy changes. There is a mention of the need to consider the heat lost by the tank water and the implications of operating under reversible conditions.
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: