# Thermodynamics: violation of 1st and/or 2nd law of thermodynamics in a heat engine

1. May 5, 2009

### rubix1225

1. The problem statement, all variables and given/known data
Here is the question:

2. Relevant equations
First law: Q=W+U
Second law: no thermodynamic system can convert heat into work 100%; heat flows from hot to cold reservoirs spontaneously; entropy always increases

3. The attempt at a solution
From the diagram, it looks like energy is conserved 100% as the heat taken from the hot reservoir (100J) is split into 75J of work and 25J carried over to the cold reservoir. Does this mean that it violates the 2nd law of thermodynamics because no system can be 100% efficient? But then again, the 2nd law heat cannot be 100% converted into WORK, and not the whole system. I'm really stumped for the 2nd law of thermodynamics in this case.

I'm not even sure what the 25J actually represents (the energy going into the cold reservoir). Does it represent the change in internal energy (U from Q=W+U)? From here, I really became stuck as to how to explain the 1st law of thermodynamics.

The answer is given as C. Could someone kindly explain why the 1st is not violated but the 2nd law is? Thanks
By the way, this question is from a past IB Physics paper and my exam is next week! Any help would be vastly appreciated.

2. May 5, 2009

### davieddy

Re: Thermodynamics: violation of 1st and/or 2nd law of thermodynamics in a heat engin

Entropy decrease of hot reservoir = 100/400 J/K
Entropy increase of cold reservoir = 25/300 J/K

Net entropy has decreased!

3. May 5, 2009

### Andrew Mason

Re: Thermodynamics: violation of 1st and/or 2nd law of thermodynamics in a heat engin

The first law is not violated because the work done by the engine is less than the net heat flow into the engine. Energy is conserved. (in one cycle, $\Delta U$ is 0, and $\Delta Q = 100 - 25 = 75 = W$).

The second law is violated for the reason given by Davieddy. You can see that the second law is violated because this engine is more efficient than a Carnot engine operating between these two temperatures. (for Carnot $\eta = 1 - Tc/Th = .25$, which is less than the efficiency here: W/Qh = .75). This is not possible by the second law.

AM

4. May 5, 2009

### davieddy

Re: Thermodynamics: violation of 1st and/or 2nd law of thermodynamics in a heat engin

I think you mean = rather than <.
THX for the "endorsement".

I taught an Andrew Mason at St Paul's School (London) around 1974,
and his whizz kid younger brother Lionel.

No relation I suppose?

David Eddy

5. May 5, 2009

### Andrew Mason

Re: Thermodynamics: violation of 1st and/or 2nd law of thermodynamics in a heat engin

I think I meant "not greater than".
Sorry, no relation. I was in London in 1974. But I don't have a younger brother Lionel - none that I know of anyway....

AM

6. May 5, 2009

### davieddy

Re: Thermodynamics: violation of 1st and/or 2nd law of thermodynamics in a heat engin

I suppose you could argue that some heat could go somewhere other
than the "cold reservoir", otherwise energy would not be "conserved".

Since your name is not uncommon, it was a bit of a long shot.
Similarly London 1974 is not an unbelievable coincidence.

7. May 5, 2009

### rubix1225

Re: Thermodynamics: violation of 1st and/or 2nd law of thermodynamics in a heat engin

Ah nonetheless, that is still quite a coincidence.
Anyway thank you for the help David and Andrew, appreciate it

8. May 6, 2009

### davieddy

Re: Thermodynamics: violation of 1st and/or 2nd law of thermodynamics in a heat engin

I thought he might be in the same time zone (time of posting) and
he has an English (as distinct from US) sounding name, and he knows

London is quite a big city, and there are quite a lot of us "babyboomers" around.

My enquiry would not have arisen if I hadn't taught someone with the same name.
I would be mildly surprised if that AM had completely forgotten me
(or at least the school) though!

But I agree, it is a nice coincidence.

David

Last edited: May 6, 2009