What is the total entropy change of a process in which 40.0 kJ of energy is transferred as heat from a large reservoir at 800. K to one at 200. K?
For 9.5, the student solutions manual uses a negative sign for (delta)S = -40000 J/800 K, and I do not know why.
9.5
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Re: 9.5
I think that the reason that it is negative is due to the fact that the first reservoir lost heat, and transferred that heat to the second reservoir, meaning that it gained heat. Losing heat is negative, gaining is positive.
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Re: 9.5
ΔS(total) = -q/T(1) + q/T(2)
The first reservoir loses heat, so the q should be negative. The second reservoir gains the heat that the first reservoir loses, so the q should be positive.
The first reservoir loses heat, so the q should be negative. The second reservoir gains the heat that the first reservoir loses, so the q should be positive.
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Re: 9.5
It is a transfer of heat, so thinking back to our ice cube in water problems, one will be positive and the other negative.
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