# 15.4 Carnot’s perfect heat engine: the second law of thermodynamics

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• Identify a Carnot cycle.
• Calculate maximum theoretical efficiency of a nuclear reactor.
• Explain how dissipative processes affect the ideal Carnot engine.

We know from the second law of thermodynamics that a heat engine cannot be 100% efficient, since there must always be some heat transfer ${Q}_{\text{c}}$ to the environment, which is often called waste heat. How efficient, then, can a heat engine be? This question was answered at a theoretical level in 1824 by a young French engineer, Sadi Carnot (1796–1832), in his study of the then-emerging heat engine technology crucial to the Industrial Revolution. He devised a theoretical cycle, now called the Carnot cycle    , which is the most efficient cyclical process possible. The second law of thermodynamics can be restated in terms of the Carnot cycle, and so what Carnot actually discovered was this fundamental law. Any heat engine employing the Carnot cycle is called a Carnot engine    .

What is crucial to the Carnot cycle—and, in fact, defines it—is that only reversible processes are used. Irreversible processes involve dissipative factors, such as friction and turbulence. This increases heat transfer ${Q}_{\text{c}}$ to the environment and reduces the efficiency of the engine. Obviously, then, reversible processes are superior.

## Carnot engine

Stated in terms of reversible processes, the second law of thermodynamics    has a third form:

A Carnot engine operating between two given temperatures has the greatest possible efficiency of any heat engine operating between these two temperatures. Furthermore, all engines employing only reversible processes have this same maximum efficiency when operating between the same given temperatures.

[link] shows the $\text{PV}$ diagram for a Carnot cycle. The cycle comprises two isothermal and two adiabatic processes. Recall that both isothermal and adiabatic processes are, in principle, reversible.

Carnot also determined the efficiency of a perfect heat engine—that is, a Carnot engine. It is always true that the efficiency of a cyclical heat engine is given by:

$\text{Eff}=\frac{{Q}_{\text{h}}-{Q}_{\text{c}}}{{Q}_{\text{h}}}=1-\frac{{Q}_{\text{c}}}{{Q}_{\text{h}}}\text{.}$

What Carnot found was that for a perfect heat engine, the ratio ${Q}_{\text{c}}/{Q}_{\text{h}}$ equals the ratio of the absolute temperatures of the heat reservoirs. That is, ${Q}_{\text{c}}/{Q}_{\text{h}}={T}_{\text{c}}/{T}_{\text{h}}$ for a Carnot engine, so that the maximum or Carnot efficiency     ${\mathrm{Eff}}_{\text{C}}$ is given by

${\mathrm{Eff}}_{\text{C}}=1-\frac{{T}_{\text{c}}}{{T}_{\text{h}}}\text{,}$

where ${T}_{\text{h}}$ and ${T}_{\text{c}}$ are in kelvins (or any other absolute temperature scale). No real heat engine can do as well as the Carnot efficiency—an actual efficiency of about 0.7 of this maximum is usually the best that can be accomplished. But the ideal Carnot engine, like the drinking bird above, while a fascinating novelty, has zero power. This makes it unrealistic for any applications.

Pls guys am having problem on these topics: latent heat of fusion, specific heat capacity and the sub topics under them.Pls who can help?
Thanks George,I appreciate.
hamidat
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Abolarin
Most especially it is the calculatory aspects that is giving me issue, but with these new strength that you guys have given me,I will put in my best to understand it again.
hamidat
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Abolarin
the distance between two suasive crests of water wave traveling of 3.6ms1 is 0.45m calculate the frequency of the wave
v=f×lemda where the velocity is given and lends also given so simply u can calculate the frequency
Abdul
You are right my brother, make frequency the subject of formula and equate the values of velocity and lamda into the equation, that all.
hamidat
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Specific heat capacity is the amount of heat required to raise the temperature of one (Kg) of a substance through one Kelvin
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i want jamb related question on this asap🙏
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