# Second Law of Thermodynamics - Entropy

1. Dec 16, 2009

### Tynged

1. The problem statement, all variables and given/known data

Premium gasoline produces 1.23×108 J of heat per gallon when it is burned at a temperature of approximately 400 ºC (although the amount can vary with the fuel mixture). If the car's engine is 25.0 % efficient, three-fourths of that heat is expelled into the air, typically at 20.0 ºC.

Part A: If your car gets 38.0 miles per gallon of gas, by how much does the car's engine change the entropy of the world when you drive 1.00 mile?

Part B: Does it decrease or increase the entropy of the world?

2. Relevant equations

$$\Delta S = S_2 - S_1 = \frac{Q}{T}$$

Where $$\Delta S$$ is the change in entropy of the system, $$S_2$$ is the entropy of the system at its final state, $$S_1$$ is the entropy of the system at its initial state, $$Q$$ is the heat added to or removed from the system, and $$T$$ is the absolute temperature at which the process is occurring.

3. The attempt at a solution

$$\Delta S = S_2 - S_1 = \frac{Q}{T} = \frac{(0.750)(1.23\times10^{8}\ J/gal)(1.00\ mi\times\frac{1\ gal}{38.0\ mi})}{673\ K} = 3610\ J/K$$

I believe I went wrong when I used the temperature of the burning fuel mixture as the absolute temperature. I am sure the temperature of the surrounding air is also important somehow. Writing the equation for entropy differently, I tried to incorporate that second temperature.

$$\Delta S = S_2 - S_1 = \frac{Q_2}{T_2} - \frac{Q_1}{T_1} = \left(\frac{(0.750)(1.23\times10^{8}\ J/gal)(1.00\ mi\times\frac{1\ gal}{38.0\ mi})}{673\ K}\right) - \left(\frac{(0.250)(1.23\times10^{8}\ J/gal)(1.00\ mi\times\frac{1\ gal}{38.0\ mi})}{293\ K}\right)$$

$$= 3610\ J/K - 2760\ J/K = 850\ J/K$$

Although this seemed to be a step in the wrong direction because my first solution was closer to the correct answer according to the automatic response.

For Part B, I assume that driving the car will increase the entropy of the world since most processes I have seen naturally tend toward increasing disorder. I am sure the correct answer to Part A will be a positive change in entropy and support my assumption.

2. Dec 17, 2009

### Andrew Mason

The total change in entropy is the change in entropy of the system + the change in entropy of the surroundings. The system keeps returning to its original state, so there is no change in the system entropy. The surroundings consist of the hot reservoir and the cold reservoir. This process can be viewed as the engine drawing heat from the hot reservoir at 400C and delivering 75% of that heat to the cold reservoir at 20C

Because heat is leaving the hot reservoir, the change in entropy of the hot reservoir is negative (dQ < 0 means dQ/T < 0). Heat is entering the cold reservoir (the air) so the entropy change of the air is positive.

$$\Delta S = \Delta S_{sys} + \Delta S_{surr} = 0 + \left(\frac{Q_h}{T_h} + \frac{Q_c}{T_c}\right) =$$