Work obtainable from two finite reservoirs

[vandergale]

The problem is as follows:

A Carnot engine operates between two finite size reservoirs, one a body of water of mass M_H at 100°C and the other a body of water of mass M_L at 0°C. Find the maximum work obtainable from the two reservoirs.

The Attempt at a Solution

I haven't done thermodynamics in a while and am trying to relearn a lot of this stuff. I first tried expressing the work done by the engine in terms of the 2nd law. I know I can write the efficiency of the engine with the ratio of the two temperatures, but I'm not sure how to find the heat removed from the hot reservoir. If someone could point me in the right direction I would be grateful.

 

[TSny]

Hello, vandergale. Welcome to PF!

Consider an infinitesimal step during the process. Corresponding to an amount of heat |dQ_H| removed from the warmer reservoir, there will be a certain amount of heat |dQ_C| added to the cooler reservoir. Can you express |dQ_C| in terms of |dQ_H| and the current temperatures T_H and T_C of the two reservoirs?

 

[Chestermiller]

Start out by drawing a PV graph of what the reversible Carnot cycle with these reservoirs looks like. Label some schematic temperature points on this graph. You need to get the "lay of the land."

 

[vandergale]

Sorry I haven't checked back, thanks you guys, its much clearer now.

 

[paranormal]

I would approach this problem by first understanding the principles of thermodynamics and how they apply to a Carnot engine. The Carnot cycle is a theoretical thermodynamic cycle that is the most efficient way to convert heat into work. In this case, the Carnot engine is operating between two finite reservoirs, which means that the amount of heat they can provide is not infinite. The efficiency of a Carnot engine is given by the equation: efficiency = 1 - (Tc/Th), where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir.

To find the maximum work obtainable from the two reservoirs, we need to consider the first and second laws of thermodynamics. The first law states that energy cannot be created or destroyed, only transferred. In this case, the energy transfer occurs as heat from the hot reservoir to the engine and then as work from the engine to the surroundings. The second law states that the total entropy of a closed system will always increase over time, or remain constant in ideal cases. This means that the heat transfer from the hot reservoir to the engine will always result in an increase in entropy.

To find the maximum work obtainable, we can use the equation: Wmax = Qh (1 - Tc/Th), where Qh is the heat transferred from the hot reservoir to the engine. To find Qh, we can use the equation Qh = MHcH (Th - Tc), where MH is the mass of the hot reservoir and cH is its specific heat capacity. Similarly, we can find the heat transferred from the cold reservoir, Qc, using the equation Qc = MLcL (Tc - Tc), where ML is the mass of the cold reservoir and cL is its specific heat capacity.

By substituting these values into the equation for maximum work, we can find the maximum work obtainable from the two finite reservoirs. It is important to note that this is the theoretical maximum and in real-world scenarios, there will always be some losses due to inefficiencies in the engine. I hope this helps to guide you in the right direction.