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 :)