As a gasoline engine is running, an amount of gasoline containing 18,500 J of chemical potential energy is burned in 1 s. During that second, the engine does 3,700 J of work. (a) What is the engine's efficiency (in percent)? % (b) The burning gasoline has a temperature of about 4,400°F (2,700 K). The waste heat from the engine flows into air at about 82°F (301 K). What is the Carnot efficiency (in percent) of a heat engine operating between these two temperatures? %

(a) Efficiency = 20%

(b) Efficiency = 88.8%

Explanation:

(a)

The efficiency of an engine is given by the following general formula:

Efficiency = Output/Input

Here, in this case:

Output = Work Done by Engine = 3700 J

Input = Chemical Potential Energy = 18500 J

Therefore,

Efficiency = 3700 J/18500 J

Efficiency = 0.2

Efficiency = 20%

(b)

the efficiency of a Carnot Engine is given by the following formula:

Efficiency = 1 - T₂/T₁

where,

T₁ = Source Temperature = 2700 K

T₂ = Sink Temperature = 301 K

Therefore,

Efficiency = 1 - 301 K/2700 K

Efficiency = 0.888

Efficiency = 88.8%

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