9.5
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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.
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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.
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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.
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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.
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