Question

20.28 Entropy Change Due to Driving. Premium gasoline produces $\mathbf{1}\mathbf{.}\mathbf{23}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}\mathbf{J}$of heat per gallon when it is burned at approximately $\mathbf{400}\mathbf{°}\mathbf{C}$(although the amount can vary with the fuel mixture). If a car’s engine is role="math" localid="1664272617143" $\mathbf{25}\mathbf{\%}$efficient, three-fourths of that heat is expelled into the air, typically at $\mathbf{20}\mathbf{°}\mathbf{C}$ . If your car gets 35 miles per gallon of gas, by how much does the car’s engine change the entropy of the world when you drive 1.0 mi ? Does it decrease or increase it?

Short answer

The car engine change the entropy of the world by $∆\mathrm{S}=3770.3\mathrm{J}/\mathrm{K}$, and since it is positive, the entropy is increasing.

Step-by-step solution

Formula for change in entropy

Heat transport divided by temperature equals the change in entropy.

Entropy change in isothermal process is given by

$\mathbf{∆}\mathbf{S}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{T}}$

Here, $∆\mathrm{S}$ is the change in entropy, $\mathrm{Q}$ is the amount of heat, and $\mathrm{T}$ is the temperature.

Given data

Heat produced by gasoline when burned at $400°\mathrm{C},\mathrm{Q}=1.23\times {10}^8\mathrm{J}/\mathrm{gallon}$,

Efficiency of car engine $=25\%$

Car gets 35 miles per gallon of gas.

Calculate entropy change at  400°C

The car burns 1 gallon gasoline for 35 miles, therefore the burned heat

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{H}}=\mathrm{Q}\times \frac{\left(1\mathrm{mi}\right)}{\left(35\mathrm{mi}/\mathrm{gallon}\right)} \\ =\left(1.23\times {10}^8\mathrm{J}/\mathrm{gallon}\right)\times \frac{\left(1\mathrm{mi}\right)}{\left(35\mathrm{mi}/\mathrm{gallon}\right)} \\ =3.5243\times {10}^6\mathrm{J} \end{gathered}$$

Therefore, the entropy change at $400°\mathrm{C}$ or $\left(673.15\mathrm{K}\right)$ is define by,

$$\begin{gathered} ∆{\mathrm{S}}_{\mathrm{H}}=\frac{{\mathrm{Q}}_{\mathrm{H}}}{{\mathrm{T}}_{\mathrm{H}}} \\ =\frac{3.5143\times {10}^6}{673.15} \\ =5220.7\mathrm{J}/\mathrm{K} \end{gathered}$$

Determine required entropy change

The car wastes the heat,

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{c}}={\mathrm{Q}}_{\mathrm{H}}·\left(1-{\mathrm{E}}_{\mathrm{f}}\right) \\ =\left(3.5143\times {10}^6\right).\left(1-\frac{25}{100}\right) \\ =2.6657\times {10}^6\mathrm{J} \end{gathered}$$

Therefore, the entropy change at the car, at role="math" localid="1664273623096" $20°\mathrm{C}\left(\mathrm{or}293.15\mathrm{K}\right)$ is define as below.

$$\begin{gathered} ∆{\mathrm{S}}_{\mathrm{C}}=\frac{{\mathrm{Q}}_{\mathrm{C}}}{{\mathrm{T}}_{\mathrm{C}}} \\ =\frac{2.6357\times {10}^6}{293.15} \\ =8991.0\mathrm{J}/\mathrm{K} \end{gathered}$$

Therefore, the entropy change is,

$$\begin{gathered} ∆\mathrm{S}=∆{\mathrm{S}}_{\mathrm{C}}-∆{\mathrm{S}}_{\mathrm{H}} \\ =\left(8991.0-5220.7\right)\mathrm{J}/\mathrm{K} \\ =3770.3\mathrm{J}/\mathrm{K} \end{gathered}$$

Hence, the entropy is increasing.