Question

As a gasoline engine is running, an amount of gasoline containing 15,000J of chemical potential energy is burned in 1 s. During that second, the engine does 3,000 J of work. (b) The burning gasoline has a temperature of about 2500 K. The waste heat from the engine flows into air at about 300 K. What is the Carnot efficiency of a heat engine operating between these two temperatures

Answer 1

`\eta_(rev) = 0.88\, (88\,\%)`,
`\eta_(real) = 0.2\,(20\,\%)`

Step-by-step explanation:

The Carnot efficiency is the maximum theoretical efficiency that a thermal machine can reach, the expression is:

`\eta_(rev) = 1 - (T_(L))/(T_(H))`

`\eta_(rev) = 1 - (300\,K)/(2500\,K)`

`\eta_(rev) = 0.88\, (88\,\%)`

The real efficiency of the engine is:

`\eta_(real) = (W)/(Q_(H))`

`\eta_(real) = (3000\,J)/(15000\,J)`

`\eta_(real) = 0.2\,(20\,\%)`

Real efficiency of the engine must be lower than maximum theoretical efficiency due to irreversibilities.