# Entropy and Gasoline

## Homework Statement

Premium gasoline produces 1.23×108 J of heat per gallon when it is burned at a temperature of approximately 400ºC (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 20.0ºC.

a) If your car gets 35.0 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?

b) Does it decrease or increase the entropy of the world?

## Homework Equations

ΔS = ΔSHOT + ΔSCOLD

## The Attempt at a Solution

a) ΔS = ΔSHOT + ΔSCOLD
ΔS = -((.75*(1.23*108 J/gal)*(1 gal/35 mile)*(1 mile))/(400+273 K) + ((.75*(1.23*108 J/gal)*(1 gal/35 mile)*(1 mile))/(20+273 K)
ΔS = -3916 J/K + 8996 J/K
ΔS = 5079 J/K

b) Entropy increases (positive value for part a)

Would both ΔSHOT and ΔSCOLD be multiplied by .75, or would one be multiplied by the .25 efficiency?

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rude man
Homework Helper
Gold Member

ΔS = ΔSHOT + ΔSCOLD

correct

## The Attempt at a Solution

a) ΔS = ΔSHOT + ΔSCOLD
ΔS = -((.75*(1.23*108 J/gal)*(1 gal/35 mile)*(1 mile))/(400+273 K) + ((.75*(1.23*108 J/gal)*(1 gal/35 mile)*(1 mile))/(20+273 K)
ΔS = -3916 J/K + 8996 J/K
ΔS = 5079 J/K

Would both ΔSHOT and ΔSCOLD be multiplied by .75, or would one be multiplied by the .25 efficiency?

What is the heat going into the engine and its temperature? What is the heat leaving the engine and its temperature?

Neither gets multiplied by 0.25.

correct

What is the heat going into the engine and its temperature? What is the heat leaving the engine and its temperature?

Neither gets multiplied by 0.25.

The heat going into the engine is 293 K and the heat leaving the engine is 673 K, so if I just multiply both by .75 (like in my attempt), will that give me the correct answer?

a) ΔS = ΔSHOT + ΔSCOLD
ΔS = ((.75*(1.23*108 J/gal)*(1 gal/35 mile)*(1 mile))/(400+273 K) - ((.75*(1.23*108 J/gal)*(1 gal/35 mile)*(1 mile))/(20+273 K)
ΔS = 3916 J/K - 8996 J/K
ΔS = -5079 J/K

So the entropy of the world decreases?

Last edited:
rude man
Homework Helper
Gold Member
The heat going into the engine is 293 K and the heat leaving the engine is 673 K, so if I just multiply both by .75 (like in my attempt), then switch the minus sign, will that give me the correct answer?

a) ΔS = ΔSHOT + ΔSCOLD
ΔS = ((.75*(1.23*108 J/gal)*(1 gal/35 mile)*(1 mile))/(400+273 K) - ((.75*(1.23*108 J/gal)*(1 gal/35 mile)*(1 mile))/(20+273 K)
ΔS = 3916 J/K - 8996 J/K
ΔS = -5079 J/K

So the entropy of the world decreases?
Why are you multiplying the heat going into the engine by 0.75? Isn't all of the heat going into the engine? Whereas, isn't some of the heat that went into the engine not going out of the engine but changed into work instead?

Oh, and I just now noticed: you have ΔS of the hot side positive and ΔS of the cold side negative. How come that?

Last edited:
Sorry, by mistake I have added.

Last edited:
Why are you multiplying the heat going into the engine by 0.75? Isn't all of the heat going into the engine? Whereas, isn't some of the heat that went into the engine not going out of the engine but changed into work instead?

Oh, and I just now noticed: you have ΔS of the hot side positive and ΔS of the cold side negative. How come that?
Oh! So, only ΔSCOLD should include .75! I switched back the signs as I had them before--I accidentally mixed them up.

ΔS = ΔSHOT + ΔSCOLD
ΔS = -((1.23*108 J/gal)*(1 gal/35 mile)*(1 mile))/(400+273 K) + ((.75*(1.23*108 J/gal)*(1 gal/35 mile)*(1 mile))/(20+273 K)
ΔS = -5221 J/K + 8996 J/K
ΔS = 3774 J/K

rude man
Homework Helper
Gold Member
That's a lot better! Everything correct now.

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