In this example, we explore the workings of a gasoline engine, focusing on the energy transformations involved in its operation. The engine takes in heat from the combustion of gasoline, performs work, and expels waste heat. The heat input, denoted as \( Q_h \), is given as \( 1.6 \times 10^4 \) joules, while the work done per cycle, \( W \), is \( 37100 \) joules. To understand the energy flow, we can visualize an energy flow diagram that illustrates the transfer of heat from a hot reservoir to a cold reservoir, with work being done in the process.

To calculate the waste heat expelled, denoted as \( Q_c \), we use the relationship for heat engines, which states that the work done is equal to the heat input minus the waste heat:

\[ W = Q_h - Q_c \]

Rearranging this equation gives us:

\[ Q_c = Q_h - W \]

Substituting the known values, we find:

\[ Q_c = 1.6 \times 10^4 \, \text{J} - 37100 \, \text{J} = 1.23 \times 10^4 \, \text{J} \]

This indicates that \( 1.23 \times 10^4 \) joules of heat is expelled as waste heat each cycle.

Next, we determine the mass of fuel burned per cycle, denoted as \( m \). The latent heat of combustion for gasoline is \( 4.6 \times 10^7 \, \text{J/kg} \). The heat input can be expressed as:

\[ Q_h = m \cdot L \]

where \( L \) is the latent heat of combustion. Rearranging for mass gives:

\[ m = \frac{Q_h}{L} \]

Substituting the values, we calculate:

\[ m = \frac{1.6 \times 10^4 \, \text{J}}{4.6 \times 10^7 \, \text{J/kg}} = 3.5 \times 10^{-4} \, \text{kg} \]

This is equivalent to approximately \( 0.35 \) grams of gasoline burned per cycle, illustrating the efficiency of gasoline engines in converting a small amount of fuel into significant work.

Finally, we calculate the power output of the engine, denoted as \( P \). The engine completes \( 60 \) cycles per second, which means the time per cycle, \( \Delta t \), is:

\[ \Delta t = \frac{1}{60} \, \text{s} \]

Using the power formula:

\[ P = \frac{W}{\Delta t} \]

we substitute the work done per cycle and the time per cycle:

\[ P = \frac{37100 \, \text{J}}{\frac{1}{60} \, \text{s}} = 2.22 \times 10^5 \, \text{W} \]

This converts to approximately \( 222 \) kilowatts, which is equivalent to about \( 297 \) horsepower, showcasing the engine's powerful output.

This example effectively illustrates the principles of thermodynamics, energy conservation, and the efficiency of heat engines, providing a practical application of the concepts learned in physics.