Solving Ohm's Law & Power Homework: Gasoline Consumption in 1 Hour

[lha08]

Homework Statement

The two headlights of a car require a total of 10 A supplied at 12 V. Given that the combustion of 1 L of gasoline releases 3X 10^7 J and that the conversion to electrical power has an efficiency of 25%, how much gasoline is consumed in one hour for this purpose alone?

Homework Equations

3. The Attempt at a Solution [/b
I'm not sure if what I did is right since I don't have the answers..but here I go:
-I first found the power by multiplying I X V...10 A X 12 V =120 W
-then I multiplied 120 W by 3600 seconds (an hour) to turn the answer into Joules...in this case 432000 Joules.
-Then I multiplied this by 0.25 to get 108 000 Joules.
-After this, I set a ratio with 1 L and 3X10^7 Joules with 108000 Joules and the unknown amount of gas. And I got 0.0036 L...
Is how I did it correct (i'm not really sure about the 25% part and the ratio part at the end...)

 

[xcvxcvvc]

Homework Statement

The two headlights of a car require a total of 10 A supplied at 12 V. Given that the combustion of 1 L of gasoline releases 3X 10^7 J and that the conversion to electrical power has an efficiency of 25%, how much gasoline is consumed in one hour for this purpose alone?

Homework Equations

3. The Attempt at a Solution [/b
I'm not sure if what I did is right since I don't have the answers..but here I go:
-I first found the power by multiplying I X V...10 A X 12 V =120 W
-then I multiplied 120 W by 3600 seconds (an hour) to turn the answer into Joules...in this case 432000 Joules.
-Then I multiplied this by 0.25 to get 108 000 Joules.
-After this, I set a ratio with 1 L and 3X10^7 Joules with 108000 Joules and the unknown amount of gas. And I got 0.0036 L...
Is how I did it correct (i'm not really sure about the 25% part and the ratio part at the end...)

You did the wrong thing when it comes to using the 25%. You found that 432KJ are needed to supply 10A at 12V for one hour. Regardless of the efficiency of the gas burning to provide this energy, that energy MUST stay the same. Otherwise, you won't have 10A drawn through the headlights with 12 V across them.

What you need to do is say .25 * gas_energy = headlights_energy. All this is saying is only 25% of the burning is actually supplying power to the headlights. The rest is lost. So all you need to do is divide by .25 instead of multiply and then finish the problem the same way you did.