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Please help, proof based on second law 
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#1
Sep1906, 10:56 AM

P: 2

Hello to everyone!
This is my first time that I use forums and I hope you will help me! Here is my question/problem: Is it possible using the second thermodynamic law to proove that: If the total work done by heat engine operated between two temperatures is zero, than it follows that the total heat absorbed by the system is greater than zero. I am sure I saw this somewhere but unfortunately I cannot found it :( Thank you in advance! 


#2
Sep2006, 03:48 AM

P: 2

Maybe I really should be more detailed in my question. So here is the whole story... Originaly I am trying to proove that the carnott heat engine is the most efficient one. To do so I take two heat engines  one carnott and the second to be any one I wish. The carnot engine is used as heat pump.
All the proofs I've seen are choosing the heat absorbed by the heat engine to be equal the heat pumped by the carnot engine and then using the second thermodynamic law they proove that the efficency of the carnott engine is the best. Till now everything is fine. But I decieded in order to proove it to choose that the work done by the heat engine is equal the work needed for the operation of the carnot heat pump. Now I need to use the same statement I am asking help for: "If W_total=0 then Q_in_total>0" In order to remove any doubt I present you my proof: " The combined heat engine consists of: #Carnot heat pump operated N cycles  W (work input), Q_2 (absobed heat from T_L), Q_1 (emitted heat to T_H) Q_1 = Q_2 + W #tested heat engine operated N' cycles  Q_in' (absorbed heat from T_H), W' (work done by the engine), Q_out (heat emitted to T_L) Q_in' = W' + Q_out' now using first summation I get: Q_in_total = Q_in'*N'  Q_1*N W_total = W'*N'  W*N at this point I decide to operate the two engines (N,N') cycles so that I get: W_total = 0 => W'*N'  W*N = 0 => W'*N' = W*N (*)  NOW I USE THE SECOND LAW IN ORDED TO SAY THAT: if W_total = 0 that Q_in_total > 0 (my question is how exactly?)  and from here it goes: Q_in_total = Q_in'*N'  Q_1*N > 0 => Q_in'*N' > Q_1*N (**) the last thing to do is compare efficiency: eta = (efficiency) = W/Q_in eta_carnot = (W)/(Q_1) = W/Q_1 = (N*W)/(N*Q_1) = {using (*)} = (N'*W')/(N*Q_1) > {using (**)} > (N'*W')/(N'*Q_in') = W'/Q_in' = eta_test therefore we get eta_carnot > eta_test " Thank you one more time! 


#3
Sep2006, 05:37 AM

Sci Advisor
HW Helper
P: 6,671

Apply the first law and use the fact that the Carnot cycle is reversible. Reversible means that when the Carnot cycle is operated in the forward direction and then reversed, the initial conditions are restored. In order for that to occur, the heat delivered to the hot reservoir on the reverse cycle is the same as the heat removed in the forward cycle. The reversal of the cycle necessarily means that the work output in the forward cycle is stored and used to drive the reverse cycle.
If there is a more efficient cycle, more heat would be delivered to the hot reservoir, but that would violate the first law. Now see if you can state the above mathematically. AM 


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