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: List the given data in consistent units.
Power $P=20$ kW, run time $t=1$ hour $=3600$ s, and efficiency $\eta=0.4$.
Step 2: Find the useful work output.
Since $1$ kW means $1$ kJ delivered each second, the useful work is $W=Pt=20\times 3600=72000$ kJ.
Step 3: Recall what efficiency means.
Thermal efficiency is the useful output divided by the heat supplied by the fuel: $\eta=\dfrac{W}{Q}$.
Step 4: Solve for the fuel energy.
Rearranging, $Q=\dfrac{W}{\eta}=\dfrac{72000}{0.4}$.
Step 5: Compute.
\[ Q=\frac{72000}{0.4}=180000\ \text{kJ} \]
Step 6: Conclude.
The fuel combustion must release $180000$ kJ of energy.
\[ \boxed{180000\ \text{kJ}} \]