Intro to thermal Physics - D. V. Schroeder -Entropy question

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Hey guys, so I am reading this book and on pages 89-90, the author says:
"Increasing temperature correspond to a decreasing slope on Entropy vs Energy graph", then a sample graph is provided, and both in that graph and in the numerical analysis given in page 87 the slope is observed to be an increasing slope too when the system increases in temperature. So I am somewhat confused.

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Chestermiller
Mentor
Hey guys, so I am reading this book and on pages 89-90, the author says:
"Increasing temperature correspond to a decreasing slope on Entropy vs Energy graph", then a sample graph is provided, and both in that graph and in the numerical analysis given in page 87 the slope is observed to be an increasing slope too when the system increases in temperature. So I am somewhat confused.
Which is the horizontal axis, energy or entropy?

robphy
Homework Helper
Gold Member
Hey guys, so I am reading this book and on pages 89-90, the author says:
"Increasing temperature correspond to a decreasing slope on Entropy vs Energy graph", then a sample graph is provided, and both in that graph and in the numerical analysis given in page 87 the slope is observed to be an increasing slope too when the system increases in temperature. So I am somewhat confused.
While the book is well-known, not everyone has it. So, it would have been good to include the graph you mention.

The horizontal axis running left-to-right [the lower axis] is the energy-count $q_A$ in the A-EinsteinSolid (where the energy is $U_A=q_Ahf$).
The horizontal axis running right-to-left [the upper axis] is the energy-count $q_B$ in the B-EinsteinSolid, where $q_A+q_B=100$ in this example.

As the energy-count $q_A$ increases (to the right),
the entropy $S_A$ is increasing (so the slope is positive)
but that slope is decreasing [since increments are decreasing]
(so temperature $T_A=\left(\frac{\partial S_A}{\partial U_A}\right)^{-1}$ is increasing).

As energy-count $q_B$ increases (to the left),
the entropy $S_B$ is increasing (so the slope is positive)
but that slope is decreasing [since increments are decreasing]
(so temperature $T_B=\left(\frac{\partial S_B}{\partial U_B}\right)^{-1}$ is increasing).

Given an initial state $(q_A,q_B)$, with $q_A+q_B=100$ for the combined system,
the system statistically evolves towards equilibrium (where $S_{total}$ has slope zero).
Here equilibrium is at $(q_A,q_B)=(60,40)$.
Since there are many more states near $(60,40)$,
if we start at $(30,70)$, the system statistically evolves to the right from $(30,70)$, and
if we start at $(70,30)$, the system statistically evolves to the left from $(70,30)$.

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