Physics 2nd law of thermodynamics question

Hi, I've been struggling to wrap my head around the 2nd law of thermodynamics and how it applies to heat engines and so was wondering if I could get some clarification.
I understand the idea that heat must flow from a hot to cold substance but not in reverse.

I think my main confusion is around this diagram and the idea that even a theoretical heat engine cannot be 100% efficient.

First of all, I was wondering what the 'source', 'sink' and the middle represent in petrol or diesel engines.
I was also wondering why Qc cannot be 0. If this were an ideal system with perfect insulation, then why would it still not be possible for alll of QH to be transferred as W.
Thank you in advance for any help!!

Hi,

Mechanical Engineering student here.

The Source is simply where you are getting energy (in the form of heat) from. Sink means where the rejected energy (in the form of heat and not converted into useful work W) goes. The bit in the middle is just a representation of the conversion process (from source energy to useful work and rejected energy). Based on this: what would you say these are in the case of internal combustion engines?

To answer your next part really just requires giving it straight:

The Second Law of Thermodynamics has a few equally valid ways of being expressed. You have mentioned one of those which is that processes proceed in a certain direction (i.e. you cannot use the 1st Law of Thermodynamics alone to determine if a process is physically possible).

Insulation is not necessarily a concern here. The answer is in relation to a property of systems called Entropy (which opens a very large kettle of fish trying to define what that is). Essentially: all processes must proceed in the direction of increasing Entropy. This is why you cannot have a work output for a system where 100% of thermal energy is converted to mechanical energy.

You can change the entropy of a system by the removal/injection of heat. So let's consider your example where no heat goes to the sink but all heat leaves the source: heat removal from the source decreases the entropy of the source. Clearly, the heat is transformed to mechanical work so there is no change in entropy in this step. And given no heat transfers to the sink at 0C then we have a situation where the entropy of the system has decreased. This is not possible.

Instead what must happen is: there must be some increase in entropy somewhere in the system larger than the decrease in entropy of the source. This is why we need a heat sink: the rejected heat increases the entropy of the sink more than the entropy of the source decreases and overall the system has an increase in entropy and thus we do not violate the 2nd Law of Thermodynamics.

Note: there are technically 3 possibilities for what can happen during a process in relation to energy.

If the entropy generation is greater than zero, we call this an irreversible process. If the entropy generation is equal to zero, we call this a reversible process. If the entropy generation is less than zero, we call this an impossible process.

Let me know what you make of this, it's quite a bit to digest.

Thank you very much for your help!

Would the source be the heat in the combustion chamber produced by the combustion reaction, I haven't really learnt about what happens in an engine outside of the 4-strokes in the combustion chamber but would the middle part represent the conversion of thermal energy in gases to work done when the gases expand and drive the piston down, and would the sink represent the wall of the piston?
Does removing heat from the source decrease its entropy because you are reducing the KE of the the molecules thus they have less disorder? And vice versa for increasing the entropy in the sink?
Also, I was wondering how it is possible that the entropy of the sink increases by a lot more than decrease in the entropy of the source if the energy loss from the source is more than the gain in energy by the sink?

Yes, that is what the source and middle part would be. The sink can really be anything else. In a 4-stroke engine, you have an intake that injects an air-fuel mixture and a route for the exhaust gases to leave via. So the rejected heat is removed by the exhaust gases the outside environment.

As an Engineering student rather than a Physics student I tend not to think of Entropy as "degree of disorder" simply because it isn't necessary for the analysis we do. The basic thing I remember is: the same gas has higher entropy at a higher temperature compared to a lower one i.e. removing heat lowers entropy and vice versa.

I'm not familiar with how to type equations in here but: a general equation that can be used (although there are details I'm intentionally missing out here) for the change in entropy of a system is that "change in entropy, deltaS = Q/T" where Q is the heat transfer for the system and T is the temperature.

As an example: say we have a source at T=800K and a sink at 500K. The source loses 2000kJ to the sink. The entropy change for the source is: -2000kJ/800K which is -2.5kJ/K. The entropy change for the sink is 2000kJ/500K which is 4kJ/K. Hence, the increase in entropy of the sink is larger than the decrease in entropy of the source and so the overall entropy of the system has increased.

A question to see if you have caught on: in the above, I assume 100% of the heat lost from the source went to the sink. Why is this okay but your scenario before of 100% of the heat becoming mechanical work and 0% going to the sink is impossible?

Thank you for the explanation, this helps a lot!
As for your question, is it because when 100% of the heat is transferred from the source to the sink, the entropy increases as your equation showed. But when 100% of heat becomes mechanical work, the entropy decreases because the entropy of the source decreases as in the equation and there is no heat transfer if all the energy becomes work, thus Q = 0.

Yes: except for the minor point that Q isn't zero, the source lost heat and hence energy. The point is that nothing gained that heat so there is no net increase in entropy. You would be correct thought to say that: Q=0 for the sink.