Question

A car engine has a power of \(20 \, kW\). The car makes a roundtrip of \(1\) hour. If the thermal efficiency of the engine is \(40\%\) and the ambient temperature is \(300 \, K\), the energy generated by fuel combustion is

• A. \(180000 \, kJ\)
• B. \(240000 \, kJ\)
• C. \(360000 \, kJ\)
• D. \(270000 \, kJ\)

Hint:

Thermal efficiency of an engine is: \[ \eta=\frac{W}{Q} \] where \(W\) is useful work output and \(Q\) is heat energy supplied by fuel.

Answer 1

The Correct Option is A

Step 1: Identify the given quantities.
Power of engine is
\[ P=20 \, kW \] Time of roundtrip is
\[ t=1 \, hour \] Convert time into seconds,
\[ t=3600 \, s \] Thermal efficiency is
\[ \eta=40\%=0.4 \]

Step 2: Calculate the useful work done by the engine.
Useful work output is given by
\[ W=Pt \] Substituting the values,
\[ W=20\times 3600 \] \[ W=72000 \, kJ \] since \(1\,kW=1\,kJ/s\).

Step 3: Use the thermal efficiency relation.
Thermal efficiency is defined as
\[ \eta=\frac{\text{Useful work output}}{\text{Heat energy supplied}} \] Let the heat energy generated by fuel combustion be \(Q\).
Then,
\[ 0.4=\frac{72000}{Q} \] \[ Q=\frac{72000}{0.4} \] \[ Q=180000 \, kJ \]

Step 4: Final conclusion.
Hence, the energy generated by fuel combustion is
\[ \boxed{180000 \, kJ} \]