Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

In a Carnot cycle, the work potential is $$W =Q_\text{in} \left(1 -\frac{T_0}{T_\text{R}}\right)$$ where $T_0$ is the temperature of the sink, $T_\text{R}$ is the temperature of the source heat reservoir, and $Q_\text{in}$ is the heat flow from the source to the sink.

What happens to the rest of the energy that is left?

That is, where does $Q_\text{in} - W $ go? What happens to it and why can't we use it?

share|cite|improve this question

2 Answers 2

Lets look at the equation first. We will rewrite the equation like this. $$W = Q_{in} - Q_{in}\dfrac{T_O}{T_R}$$

In the limit where $$\lim_{T_R\to\infty}\dfrac{T_O}{T_R} = 0$$

We can observe that $$W = Q_{in}$$

and when $T_O = T_R$ then $$W = Q_{in} - Q_{in} = 0$$

Which highlights the ratio as a type of hyperbolic phase factor that determines the amount of work that can be extracted from an amount of heat based on the temperature reservoir compared to some ground state temperature.

When we discuss the Carnot cycle, we need to introduce the concept of entropy, and see that work out of the system is an area bounded by a curve in the T-S plane (which for simplicity is usually shown as a square). This situation can be described with equation for work as:

$$W=(T_R - T_O)(S_B - S_A) = \Delta T\Delta S$$

This equation makes it clear that work can only be extracted when there is a change in entropy which is the thermal energy per unit temperature unavailable for work. So the temperature ratio is a factor that informs us how much of the input heat is unavailable for work. What entropy is was not well defined for many years until Boltzmann was able to provide a microscopic interpretation.

Boltzmann understood entropy as being related to the number of configurations atoms could find themselves in. When there are more possible configurations, the entropy is high. What happens in the real world is that the number of possible configurations associated with the background is enormous, so while some heat is converted to work, another portion of the heat is absorbed by the environment to energize some of the environmental configurations. It is essentially a tax we pay in order to do useful work. What is really happening is that the system is trying to reach equilibrium, and in fact the heat that isn't used for work is actually heating the environment directly (and in small closed system, the environment would actually be increasing in temp, we don't notice the effect because the environment is so large in comparison to the systems we deal with).

share|cite|improve this answer

The rest of the heat, $Q_{out}=Q_{in}-W$ flows into the sink at $T_0$. It is a consequence of the entropy principle (total entropy never decreases) that $Q_{out}$ cannot be less than ($Q_{in} -$ the calculated work potential). (It could be greater, in which case the work done would be less than the work potential.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.