I have a doubt over the Kelvin and Planck's statement of thermodynamics' second law, in particular applied to a cycle. Let's take a Carnot cycle as an example, and let's call the first two transformations (the isotherm and the adiabatic) done. Now, isn't it obvious that the machine has to give up heat to go back to the initial state? Isn't it something that follows from the fact that the cycle has to be continuous?
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asked Jun 27, 2013 at 17:55
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1$\begingroup$ Can you explicitly write the statement of the second law you are using? $\endgroup$– joshphysicsJun 27, 2013 at 18:04
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$\begingroup$ Sure. "It's impossible to realise a process which has as the sole result the transformation in work of the heat given by a reservoir at a uniform temperature" $\endgroup$– Gennaro Marco DevincenzisJun 27, 2013 at 18:14
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$\begingroup$ Right so in the Carnot cycle, the engine takes some heat $|Q_H|$ from a hot reservoir, and transforms some of it into work $W$ and the rest into exhaust heat $|Q_L|$ which is given up to a lower temperature reservoir. The exhaust heat is precisely what causes the law to not be violated. $\endgroup$– joshphysicsJun 27, 2013 at 18:20
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$\begingroup$ Thank you. My doubt was just that a general cycle, once has done work, has to close itself in some way (I'm thinking of it in a p-V diagram), and I couldn't think of any plausible way in which a cycle could close without giving up some exhaust heat. The engine would have to "jump" in some manner to go back to the initial state. But as I understood from your comment, that's precisely what the Kelvin and Planck's statement says, right? $\endgroup$– Gennaro Marco DevincenzisJun 27, 2013 at 18:39
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$\begingroup$ Yeah that's basically right. Any cycle is going to have to give up some exhaust heat to satisfy the Kelvin statement. I'm going to make this a bit more precise in an answer. $\endgroup$– joshphysicsJun 27, 2013 at 18:41
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2 Answers
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Every thermodynamic system satisfies the first law during a given process; $$ \Delta E = Q-W $$ Here $\Delta E$ is the change in its internal energy, $Q$ is the heat transferred to the system, and $W$ is the work done by the system. For a system undergoing a cyclic process, namely one for which it starts and ends in the same thermodynamic state, one has $\Delta E = 0$, and the first law then tells us that $$ Q = W $$ Now, suppose that the system is taking some heat $Q_H>0$ from a reservoir, and turning it into some work $W$. Let's define $$ Q_\mathrm{exhuast} = Q-Q_H $$ then we can write $$ Q_H + Q_\mathrm{exhaust} = W $$ The Kelvin statement tells us that we must have $Q_\mathrm{exhaust}\neq 0$ because otherwise, we would have $Q_H = W$; the sole result of the process would have been to transform the heat it took in into work. The symbol $Q_\mathrm{exhaust}$ is what one commonly calls the exhaust heat.
answered Jun 27, 2013 at 18:54
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$\begingroup$ Thank you very very much. Too bad I can't give you reputation. :( $\endgroup$ Jun 27, 2013 at 18:59
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in simple words we can expain kp statement as it is impossible to completely convert heat into work in a cycle(*).
while in case of a single process say a isothermal expansion as dt=0 therfore ineternal energy which is a function of temp so dE = 0 neglecting k.e and p.e so from 1st law we get dq = dw .
in isothermal expansion we are able to get work but only till the pressure of the gas do not becomes equal to the atmospheric pressure in which we can convert heat into work but not infinitely.
hence the kp statement is that we cannot completely convert heat into work in a cycle of infinitely.....
