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- Homework Statement
- Three blocks of metal have the same mass and specific heat. Their initial temperatures are T, 2T, and 3T. Reversible heat engines are connected between these blocks, but no net work is produced or used. No heat is exchanged with the surroundings. Finally two of the blocks reach the same temperature T1, and the third block reaches T2. Find these temperatures.

- Relevant Equations
- First law of thermodynamics

Second law of thermodynamics

Hi,

I posted a similar question recently and gained some insight on these types of problems. However, I am slightly stumped on how to approach this variation of the problem.

So I know that:

- there is no net change in enthalpy of the blocks and the engine as the processes are reversible

- [itex] \Delta S_{surroundings} = 0 [/itex] as no heat is exchanged with surroundings

- the internal energy of the system should not change as it is a cycle

I have made a quick sketch of what I envision system to look like - but perhaps I am wrong. I have assumed that the work extracted from a certain turbine connecting block i to j is used to power the pump between those same blocks.

(have left out work terms for simplicity in the drawing)

I tried to start writing equations for [itex] dQ_i [/itex] for each block, but am ending up with lots of big equations (am thinking that surely there is a more elegant way of solving this). For example, if we number the blocks 1, 2, and 3 (with the numbering corresponding to the blocks with the initial termperature as [itex] nT [/itex]), then I am getting the following equation for [itex] dQ_1 [/itex]:

[tex] dQ_1 = mcdT_2 - dW_{21} - mcdT_1 - mcdT_1 + mcdT_3 + dW_{31} [/tex]

doing the same for [itex] dQ_2 [/itex] and [itex] dQ_3 [/itex], I get:

[tex] dQ_2 = mcdT_3 - dW_{23} - 2mcdT_2 + mcdT_1 + dW_{21} [/tex]

[tex] dQ_3 = mcdT_2 + dW_{23} - 2mcdT_3 + mcdT_1 + dW_{31} [/tex]

Then, given that we know that [itex] dS = \frac{dQ}{T} = 0 [/itex], then combining those equations, I get:

[tex] \frac{dQ_1}{T_1} + \frac{dQ_2}{T_2} + \frac{dQ_3}{T_3} = 0 [/tex]

rearranging gives:

[tex] mcdT_1 \left( \frac{1}{T_2} + \frac{1}{T_3} + \frac{-2}{T_1} \right) + mcdT_2 \left( \frac{1}{T_3} + \frac{1}{T_1} + \frac{-2}{T_2} \right) + mcdT_3 \left( \frac{1}{T_2} + \frac{1}{T_1} + \frac{-2}{T_3} \right) + dW_{13} \left( \frac{1}{T_3} - \frac{1}{T_1} \right) + dW_{12} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) + dW_{23} \left( \frac{1}{T_3} - \frac{1}{T_2} \right) = 0 [/tex]

I am not really sure if what to do from here (or if this was the right path to follow). I could integrate, but then I am not sure what to do with all the work terms that I have.

Any help would be greatly appreciated.

I posted a similar question recently and gained some insight on these types of problems. However, I am slightly stumped on how to approach this variation of the problem.

So I know that:

- there is no net change in enthalpy of the blocks and the engine as the processes are reversible

- [itex] \Delta S_{surroundings} = 0 [/itex] as no heat is exchanged with surroundings

- the internal energy of the system should not change as it is a cycle

I have made a quick sketch of what I envision system to look like - but perhaps I am wrong. I have assumed that the work extracted from a certain turbine connecting block i to j is used to power the pump between those same blocks.

(have left out work terms for simplicity in the drawing)

I tried to start writing equations for [itex] dQ_i [/itex] for each block, but am ending up with lots of big equations (am thinking that surely there is a more elegant way of solving this). For example, if we number the blocks 1, 2, and 3 (with the numbering corresponding to the blocks with the initial termperature as [itex] nT [/itex]), then I am getting the following equation for [itex] dQ_1 [/itex]:

[tex] dQ_1 = mcdT_2 - dW_{21} - mcdT_1 - mcdT_1 + mcdT_3 + dW_{31} [/tex]

doing the same for [itex] dQ_2 [/itex] and [itex] dQ_3 [/itex], I get:

[tex] dQ_2 = mcdT_3 - dW_{23} - 2mcdT_2 + mcdT_1 + dW_{21} [/tex]

[tex] dQ_3 = mcdT_2 + dW_{23} - 2mcdT_3 + mcdT_1 + dW_{31} [/tex]

Then, given that we know that [itex] dS = \frac{dQ}{T} = 0 [/itex], then combining those equations, I get:

[tex] \frac{dQ_1}{T_1} + \frac{dQ_2}{T_2} + \frac{dQ_3}{T_3} = 0 [/tex]

rearranging gives:

[tex] mcdT_1 \left( \frac{1}{T_2} + \frac{1}{T_3} + \frac{-2}{T_1} \right) + mcdT_2 \left( \frac{1}{T_3} + \frac{1}{T_1} + \frac{-2}{T_2} \right) + mcdT_3 \left( \frac{1}{T_2} + \frac{1}{T_1} + \frac{-2}{T_3} \right) + dW_{13} \left( \frac{1}{T_3} - \frac{1}{T_1} \right) + dW_{12} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) + dW_{23} \left( \frac{1}{T_3} - \frac{1}{T_2} \right) = 0 [/tex]

I am not really sure if what to do from here (or if this was the right path to follow). I could integrate, but then I am not sure what to do with all the work terms that I have.

Any help would be greatly appreciated.

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