9.5

$\Delta S = \frac{q_{rev}}{T}$

Annie Lieu-1H
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9.5

Why is q/t for 800 K negative but q/t for 200k is positive?

Meredith Steinberg 2E
Posts: 50
Joined: Thu Jul 13, 2017 3:00 am

Re: 9.5

Heat is leaving the large reservoir at 800K, which explains why the entropy is negative.

Lindsay Kester 2L
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Joined: Thu Jul 27, 2017 3:00 am

Re: 9.5

Whenever heat is leaving something, the amount of possible spaces (W) that the molecules can take up decreases, so entropy does as well.

Nehal Banik
Posts: 64
Joined: Thu Jul 13, 2017 3:00 am

Re: 9.5

Since it is at 800K, the energy is leaving the system therefore that energy is negative, whereas the surroundings absorbs energy therefore it is positive, therefore it makes sense to put the negative there.

Johann Park 2B
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Re: 9.5

∆STOT = ∆SSYS + ∆SSURR. When heat is leaving the system, the value is negative. As heat leaves the system, it enters the surroundings.

Liz White 1K
Posts: 35
Joined: Thu Jul 13, 2017 3:00 am

Re: 9.5

In addition, when using the equation deltaS = qrev/T, you'll find that a lower temperature (in the denominator) will result in a greater overall entropy deltaS, and vice versa. So, 200K will result in a greater deltaS (deltaS = 200) than 800K (deltaS = 50). They are inversely proportional. To solve for the difference in this equation, you just do final - inital (200-50), which results in an entropy change of 150 J/K.