Reference: Daniel V. Schroeder, *An Introduction to Thermal Physics*, (Addison-Wesley, 2000) – Problems 3.10 – 3.11.

The original definition of entropy was in terms of heat flow, rather than the multiplicity of states in a system. If an amount of heat flows into a substance at a constant temperature , the change in entropy is

If heat flows *out* of the substance, then is negative and the system loses entropy. In order for a system to maintain a constant temperature when it gains or loses heat, it must differ from the types of systems we’ve considered up to now. One possibility is that the amount of heat gained is very small compared to the existing internal energy of the system so that the added heat makes a negligible difference to the temperature. Such a system is called a *heat reservoir*. Another example is a phase change, as when ice melts into liquid water, as during a phase change, heat is gained or lost and the substance doesn’t change its temperature.

Example 1Suppose a 30 g ice cube at is placed on a table in a room at . The ice will first melt into water, still at 273 K (because it’s a phase change), then the water will warm up to 298 K. All of this heat is transferred from the air in the room, which we can consider to be a heat reservoir at a constant temperature of 298 K. The changes in entropy are then:The latent heat of fusion of water at 273 K is , so the amount of heat required to melt the ice is

Since it occurs at a constant temperature, the entropy change of the water is

As the water warms up, it absorbs heat, but the temperature varies. We can then use our relation between entropy and heat capacity, along with the fact that the specific heat capacity of water is roughly constant over the liquid range at to get

Thus the total entropy increase of the water is

The total amount of heat transferred to the water from the room’s air is

This happens at a constant room temperature of 298 K so the entropy lost by the room is

The net entropy change of the universe is therefore

which is positive, as required by the second law of thermodynamics.

Example 2To draw a bath (in the days before hot and cold running water, presumably) we mix 50 litres of hot water at with 25 litres of cold water at . The final temperature of the water is the weighted average:To find the entropy change, we can use 5. The hot water cools down and thus loses heat, so its entropy change is

The entropy gained by the cold water is

Thus the net entropy change is

which is again positive.

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Tabitha Booth-SeayOn line 9, wouldn’t the Q= -(334 J/g * 30 g + 30 g * 4.184 J/gK * (298-273)) = -13155.8 J

making Sroom = -13115.8 J / 298 K = -44.1 J/K ?

Without adding in the temp change on line 9, the units would not work out properly.

gwrowePost authorFixed now. Thanks.