# Change in entropy of the universe due to a carnot heat engine

If you consider a Carnot engine operating between a hot reservoir, $$T_b$$ and a cold reservoir $$T_s$$. I was asked to find an expression for the total change in entropy of the universe (part D).

Exact details:

I am not sure how I would go about this. The only fact I know is that the change in entropy over 1 cycle for the Carnot engine is 0.

I said that based on the second law of thermodynamics:

$$\Delta S \geqslant 0.$$

$$\Delta S_{universe}+\Delta S_{T_s \ res} + \Delta S_{T_b \ res} + \Delta S_{probe} \geqslant 0$$ (note the probe being the carnot engine)

I worked out the that the change in entropy of the hot and cold reservoir cancels, and by saying that the change in entropy of a Carnot cycle is 0, I get the obvious result that:

$$\Delta S_{universe} \geqslant 0$$.

• Not sure what's going on here, but if the probe is truly operating in a Carnot cycle, then $\Delta S_{universe}=0$. – Bob D Jan 3 at 14:38
• Essentially, there is a power source which is emitting heat to the probe ( hot reservoir giving heat to the engine ) and the surface where the probe is rejecting heat to ( cold reservoir ). I have shown based on the given equation for power generated by the nuclear decay, the total work will be that given in part C. I just dont understand why there will be a net change in entropy. – Vishal Jain Jan 3 at 14:52
• If the engine does not operate reversibly, even though the change in entropy of the engine working fluid in a cycle is zero, the heat exchanges with the combination of reservoirs is such that their combined entropy increases. – Chet Miller Jan 3 at 16:14
• It says to assume an ideal carnot engine, which should operate reversibly, If I'm not mistaken. – Vishal Jain Jan 3 at 16:32
• Then the equal sign applies. – Chet Miller Jan 3 at 16:42

I can see what is happening here. Entropy is being generated within the block of radioactive material as a result of the radioactive decay reaction. What they expect you to assume is that this generated entropy is all exactly removed from the block by transferring heat to the engine. So the entropy change of the block is assumed to be exactly zero.

If we let $$Q_H$$ represent the total heat transferred from the block to the engine and $$Q_C$$ represent the total heat transferred from the engine to the cold reservoir, then the changes in entropy of the hot block, the engine, and cold reservoir are:

$$\Delta S_{block}=-\frac{Q_H}{T_b}+\sigma_b=-\frac{\int_0^{\infty}{P_{in}(t)dt}}{T_b}+\sigma_b=-\frac{P_0\tau}{T_b}+\sigma_b=0$$ $$\Delta S_{engine}=\frac{Q_H}{T_b}-\frac{Q_C}{T_s}=0$$ and $$\Delta S_{cold}=+\frac{Q_C}{T_s}$$where $$\sigma_b$$ is the total entropy generated within the block as a result of nuclear reaction. If we add there three entropy changes together, we obtain the entropy change of the universe: $$\Delta S_{universe}=\Delta S_{block}+\Delta S_{engine}+\Delta S_{cold}=\frac{Q_C}{T_s}=\frac{Q_H}{T_b}=\sigma_b=\frac{P_0\tau}{T_b}$$

So this all depends on the assumption that the entropy removed from the hot block is exactly compensated by the entropy generated by nuclear reaction in the hot block, such that the change in entropy of the hot block is zero.

• Thanks this explains it perfectly, just one question in regards to why we dont say that the change in entropy of the universe is greater than or equal to $\frac{P_0\tau}{T_b}$. Why is it strictly equal to the change? – Vishal Jain Jan 4 at 8:19
• Also I tried to solve the question based on your answer but I came across the following issue. I worked from this statement: $\Delta S_{universe}+\Delta S_{block}+\Delta S_{engine}+\Delta S_{cold}=0$ Where all the changes in entropies are the same as in your answer. But then this leads me to have: $\Delta S_{universe}=-\Delta S_{block}-\Delta S_{engine}-\Delta S_{cold}=-\sigma_b=-\frac{P_0\tau}{T_b}$ This is the negative of the correct answer, why is the initial statement wrong? – Vishal Jain Jan 4 at 8:58
• The Clausius inequality, when applied properly to the universe (in this case, the adiabatic grouping of block, engine, and cold sink) requires that $$\Delta S_{universe}\geq 0$$ It is certainly satisfied by $$\Delta S_{universe}=\frac{P_0\tau}{T_b}\geq0$$ – Chet Miller Jan 4 at 13:05
• Your initial statement $$\Delta S_{universe}+\Delta S_{block}+\Delta S_{engine}+\Delta S_{cold}=0$$makes no sense to me whatsoever. I have no idea what your motivation was for writing this. The correct equation should have been $$\Delta S_{universe}=\Delta S_{block}+\Delta S_{engine}+\Delta S_{cold}+\Delta S_{everything\ else}$$But in this problem $\Delta S_{everything\ else}=0$ since the block, engine, and cold sink are isolated thermally from everything else and only adiabatic reversible work is done on everything else. – Chet Miller Jan 4 at 13:15
• I was thinking about the statement $\Delta S_{process} \geqslant 0$ And so I just summed the change in entropies on the LHS and equated it to 0 assuming it all cancels out in a reversible process, my mistake was thinking of $\Delta S_{everything \ else}$ as $\Delta S_{universe}$. – Vishal Jain Jan 5 at 13:32