Understanding Heat Engines: Exploring the Second Law of Thermodynamics

[Coop]

Hi,

I am trying to figure out heat engines. I don't understand why heat from a hot reservoir MUST exhaust heat into a cold reservoir. How does that satisfy the second law of thermodynamics? Why can't all energy from the hot reservoir be used to do work?

Thanks

 

[dauto]

If all the heat from the hot reservoir is used to produce work and that work is used to run a heat pump that pumps heat from a cold reservoir back to the hot reservoir the hot reservoir gets even hotter and you created a perpetual motion machine of the second kind which is forbidden by the second law of thermodynamics.

 

[Coop]

Thanks :) Can you explain why putting all the energy from the hot reservoir to work would cause entropy of the system to decrease?

 

[dauto]

Thanks :) Can you explain why putting all the energy from the hot reservoir to work would cause entropy of the system to decrease?

The formula for the change in entropy is $dS=\frac{\delta Q}{T}$, where $\delta Q$ is the heat lost and is negative while $T$ is the temperature and is positive. Clearly $dS$ is negative representing a decrease of entropy forbidden by the 2nd. If some of the heat goes to a second reservoir at lower temperature than there is a second term in the expression for the entropy. $dS=\frac{\delta Q_1}{T}_1 + \frac{\delta Q_2}{T}_2$, $\delta Q_1$ is the heat lost at the higher temperature reservoir $T_1$ and is still negative, but $\delta Q_2$ is the heat gained at the lower temperature reservoir $T_2$ and is positive. Since $T_2$ is smaller than $T_1$ the second term can counter the first leading to a positive increase of entropy and there will be some heat (but not all) left over to produce work.

 

[Coop]

The formula for the change in entropy is $dS=\frac{\delta Q}{T}$, where $\delta Q$ is the heat lost and is negative while $T$ is the temperature and is positive. Clearly $dS$ is negative representing a decrease of entropy forbidden by the 2nd. If some of the heat goes to a second reservoir at lower temperature than there is a second term in the expression for the entropy. $dS=\frac{\delta Q_1}{T}_1 + \frac{\delta Q_2}{T}_2$, $\delta Q_1$ is the heat lost at the higher temperature reservoir $T_1$ and is still negative, but $\delta Q_2$ is the heat gained at the lower temperature reservoir $T_2$ and is positive. Since $T_2$ is smaller than $T_1$ the second term can counter the first leading to a positive increase of entropy and there will be some heat (but not all) left over to produce work.

Thank you :)