Q3E - University Physics with Modern Physics

Question

Question: A Gasoline Engine. A gasoline engine takes in \(1.61 \times 1{0^4}\;J\) of heat and delivers \(3700\;J\) of work per cycle. The heat is obtained by burning gasoline with a heat of combustion of  (a) What is the thermal efficiency? (b) How much heat is discarded in each cycle? (c) What mass of fuel is burned in each cycle? (d) If the engine goes through 60.0 cycles per second, what is its power output in kilowatts? In horsepower? 

Step-by-Step Solution

Answer

The thermal efficiency the gasoline engine is \(28.88\% \).

Step 1: Write the given data from the question.

Gasoline engine takes amount of heat,\({Q_H} = 1.61 \times {10^4}\;{\rm{J}}\) 

Mechanical work output,\(W = 3700\;{\rm{J}}\) 

Step 2: Determine the formulas to calculate the thermal efficiency of the engine.

The thermal heat energy of the engine is defined as the amount of the heat that is converted into work done.

The expression to calculate the thermal efficiency of the gasoline engine is given as follows.

\({\eta _{th}} = \frac{W}{{{Q_H}}}\)                                                                       …… (i)

Here, \(W\) is the mechanical work and \({Q_H}\) is the heat that must be supplied to diesel engine.

Step 3: Calculate the thermal efficiency of the engine.

Calculate the thermal efficiency of the gasoline engine.

Substitute \(3700\;{\rm{J}}\) for \(W\) and \(1.61 \times {10^4}\;{\rm{J}}\) for \({Q_H}\) into equation (i).

\(\begin{array}{l}{\eta _{th}} = \frac{{3700}}{{1.61 \times {{10}^4}}} \times 100\\{\eta _{th}} = \frac{{37}}{{1.61}}\\{\eta _{th}} = \frac{{3700}}{{161}}\\{\eta _{th}} = 22.98\% \end{array}\) 

Hence the thermal efficiency the gasoline engine is \(28.88\% \).