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Entropy change and maximum work problem

Consider a thermally isolated system that consists of a body at temperature T1 and a large reservoir at temperature T2 > T1. The reservoir is sufficiently large so that its temperature does not change. The body has constant pressure heat capacity of Cp that does not vary with temperature.

Please explain to me how to obtain the change in entropy of the universe of this system.

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Solution Preview

Because the problem is rather simple, I included an extra problem about the maximum amount of work that can be extracted from this process.

Due to heat flow from the reservoir and the object, the temperature of the object will increase. The amount of heat that flows to the object can be expressed in terms of the temperature increase of the object as:

dq = Cp dT1 (1)

When dq of heat flows from the reservoir to the object, the entropy of the reservoir decreases, we have:

dS2 = -dq/T2 (2)

where S2 is the entropy of the reservoir.

The entropy of the body, S1, changes according to:

dS1 = dq/T1 (3)

The change in the total entropy S is thus:

dS = dS1 + dS2 = dq (1/T1 - 1/T2) (4)

Using (1) we can express this in terms of the temperature change of the object:

dS = Cp dT1 (1/T1 - 1/T2)

If we start with the object at some temperature T1 < T2, we'll eventually end up with the object at the same ...

Solution Summary

The solution explains how to compute the entropy change. It also explains how the formula for the entropy increase can be used to obtain the expression for the maximum amount of work that can be extracted from the system without doing any additional computations. This despite that process being different as the process is now reversible.