Efficiency of a cycle from a diagram

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The discussion focuses on calculating the efficiency of a heat machine using a Carnot cycle diagram. The user struggles with evaluating the heat transfer (\Delta Q) for the first process in the cycle, initially misapplying the equations. After receiving guidance on the correct integration of temperature and entropy, they successfully arrive at the same efficiency formula as a peer. The conversation also hints at further exploration of the Carnot cycle, indicating a willingness to seek additional help if needed. Overall, the thread emphasizes the importance of proper application of thermodynamic principles in calculating efficiency.
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Homework Statement


Calculate the efficiency of the heat machine that is shown in the figure. Draw the corresponding Carnot cycle diagram in the S-T plane.


Homework Equations


\varepsilon = \frac{ \Delta W}{\Delta Q}.
\Delta Q = T \Delta S

The Attempt at a Solution


My problem lies in evaluating \Delta Q for the process 1 (or from A to B).
For the process 2, \Delta Q=T_2(S_2-S_1).
For the process 3, \Delta Q=0 because there's no change in entropy.
But I'm stuck at process 1. My attempt was \Delta Q=\Delta T \Delta S=(T_2-T_1)(S_2-S_1) but I know this doesn't make any sense.
Using that non sensical result I find \varepsilon = \frac{1}{T_2-\frac{T_2^2}{T_1}} which makes no sense because if T_1=T_2 I get an infinite efficiency while I should get 0. A friend of mine reached \varepsilon =\frac{T_1-T_2}{T_1+T_2} but I don't know how he did nor if that's right either.
Any help is appreciated.
 

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Did you consider that ##dQ=TdS##, and that if T changes with S that then ##ΔQ=∫TdS##?
 
I like Serena said:
Did you consider that ##dQ=TdS##, and that if T changes with S that then ##ΔQ=∫TdS##?
Actually not! haha.
Now I reach the same answer as my friend. Thank you very much for that. :biggrin:
Hmm I'll think about the Carnot cycle. If I need help I'll post here.
 
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