# Second Law of Thermodynamics - Entropy

• Tynged
In summary, the car's engine, which is 25.0% efficient, expels three-fourths of the 1.23×10^8 J of heat produced by burning premium gasoline into the air at 20.0 ºC. To calculate the change in entropy of the world when driving 1.00 mile with a car that gets 38.0 miles per gallon, the equation \Delta S = S_2 - S_1 = \frac{Q}{T} is used. The answer is 3610 J/K, which is a positive change in entropy, supporting the assumption that driving the car increases the entropy of the world. However, the correct calculation includes the temperature of the hot and cold
Tynged

## Homework Statement

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?

## Homework 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.

## 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.

Tynged said:

## 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.
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.

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

AM

Your approach to solving for the change in entropy is correct. However, the absolute temperature used in the calculation should be the average temperature of the system, which in this case is the average of the burning temperature (400 ºC) and the surrounding air temperature (20 ºC). So the correct absolute temperature to use is 240 ºC or 513 K. This will give a change in entropy of 2420 J/K, which is closer to the answer provided.

For Part B, you are correct in assuming that driving the car will increase the entropy of the world. This is because the process of burning gasoline and converting it into mechanical energy is not 100% efficient, resulting in the production of waste heat that is expelled into the surrounding environment. This waste heat disperses and increases the overall disorder of the system, which is a characteristic of the second law of thermodynamics. Therefore, driving the car does contribute to the increase in entropy of the world.

## What is the Second Law of Thermodynamics?

The Second Law of Thermodynamics states that the total entropy of an isolated system will always increase over time. In other words, the natural tendency of energy and matter is to become more disordered and spread out.

## What is entropy?

Entropy is a measure of the disorder or randomness in a system. The higher the entropy, the more disordered the system is. It is a concept used in thermodynamics to explain the direction of energy flow and the tendency towards equilibrium.

## How does the Second Law of Thermodynamics relate to the concept of energy conservation?

The Second Law of Thermodynamics does not contradict the law of energy conservation. Energy is always conserved, but it can change forms and become less useful or more dispersed. The Second Law explains how energy changes from a concentrated, organized form to a more dispersed, disorganized form.

## Can entropy ever decrease in a system?

According to the Second Law of Thermodynamics, the total entropy of an isolated system will always increase. However, in a non-isolated system, such as living organisms, entropy can decrease locally by using energy to create order. This decrease in entropy is offset by an overall increase in the entropy of the surrounding system.

## What are some real-world examples of the Second Law of Thermodynamics?

Some common examples of the Second Law of Thermodynamics include the cooling of a hot cup of coffee, the rusting of metal, and the decay of organic matter. In all of these cases, energy is being dispersed and entropy is increasing over time.

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