i need help with this physics

Question

As a gasoline engine is running, an amount of gasoline containing 12,000 J of chemical potential energy is burned in 1 s. During that second, the engine does 4,000 J of work.

(a)

What is the engine's efficiency (in percent)?

 %

(b)

The burning gasoline has a temperature of about 5,000°F (3,000 K). The waste heat from the engine flows into air at about 88°F (304 K). What is the Carnot efficiency (in percent) of a heat engine operating between these two temperatures?

 %

Answer 1

Engine efficiency = power the engine outputs / power put into the engine. Power is just work over time, and time is 1s. So we have 4,000J/1s = 4000W out, and 12,000J/1s = 12,000W in.

This gives us 4000W/12,000W, which equals 0.333 or 33.3% (This number might seem low, but anything between 30-40% for a gas-powered engine is pretty good)

Carnot efficiency is the theoretical maximum efficiency of a heat engine. The equation you use to find it is:

max efficiency = 1 - (Tl/Th), where Tl is the temperature of the cold sink (air), and Th is the temperature of the hot sink (burning gasoline).

We get 1 - (304K/3000K) = 0.899 or 89.9%.

The theoretical maximum should always be higher than the actual amount. So to do a quick error check for problems like these, 89.9% should be > 33.3%. And they are, so things look good!