Why Does Entropy Behave Differently in Water and Steam with Temperature Changes?

AI Thread Summary
The discussion centers on the behavior of entropy in water and steam as temperature changes. It is noted that the entropy of water increases with temperature due to the energy required for phase transition to steam at 100 degrees Celsius. In contrast, the entropy of steam appears to decrease with temperature when considering specific conditions, such as increasing pressure, which is clarified through steam tables. The conversation emphasizes the importance of defining conditions, particularly pressure, when discussing entropy changes. Overall, the relationship between temperature, pressure, and entropy in both phases is complex and requires careful consideration.
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Homework Statement



Why does the entropy of the water increase with increasing temperature, while the entropy of steam decreases, while the entropy of steam decreases with increasing temperature?

The Attempt at a Solution



I think the reason why the entropy increases with water is because it takes an extra amount of heat to convert water to steam when it reaches a temperature of 100 degrees celsius more than increasng the temperature of water while since when increasing the temperature of steam , its not gong through a phase transition and so as the temperature goes up, the entropy goes down since S=Q/T
 
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I'm stuck at the premise of the question. Who says that the entropy of steam decreases with increasing temperature? That would mean that the specific heat of steam is negative, which is untrue.
 
If you look at the steam tables, you will find that the entropy of steam in the saturated vapor (steam) condition does decrease as the temperature and pressure both increase.

At a pressure of 0.6113 kPa and a saturation temperature of 0.01 C the entropy is 9.1562 kJ/kg-K.

At the critical point of 22.09 MPa and a saturation temperature of 374.14 C the entropy is 4.4298 kJ/kg-k

Now, a saturated liquid (water) has an increase in entropy as the temperature and pressure both increase.

So the original question is justified as long as one indicates that the pressure is increasing also.

Thanks
Matt
 
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Ah, got it. I was thinking (\partial S/\partial T)_P or (\partial S/\partial T)_V. This would be more like (\partial S/\partial T)_{\mu=\mu(H_2O(l),\,T,\,P)}
 
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