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

Answer:

$\eta_{rev} = 0.88\, (88\,\%)$, $\eta_{real} = 0.2\,(20\,\%)$

Explanation:

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

$\eta_{rev} = 1 - \frac{T_{L}}{T_{H}}$

$\eta_{rev} = 1 - \frac{300\,K}{2500\,K}$

$\eta_{rev} = 0.88\, (88\,\%)$

The real efficiency of the engine is:

$\eta_{real} = \frac{W}{Q_{H}}$

$\eta_{real} = \frac{3000\,J}{15000\,J}$

$\eta_{real} = 0.2\,(20\,\%)$

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