The question appears to involve calculating the thermal efficiency of a car engine based on the given data: - Fuel consumption rate: 22.4 L/hr - Power output: 55 kW - Heating value of fuel: 40,000 kJ/kg - Density of fuel: .8 g/cm³ **Step-by-step solution:** ### 1. Convert fuel consumption to mass flow rate - Volume flow rate: 22.4 L/hr - 1 L = 100 cm³ - Volume per hour: \( 22.4 \times 100 = 22400 \text{ cm}^3/\text{hr} \) - Convert to mL: 22400 mL/hr (since 1 cm³ = 1 mL) - Density: .8 g/cm³ - Mass flow rate: \[ \text{Mass} = \text{Volume} \times \text{Density} = 22400 \text{ cm}^3/hr \times .8 \text{ g/cm}^3 = 17920 \text{ g/hr} \] - Convert to kg/hr: \[ 17920 \text{ g/hr} \div 100 = 17.92 \text{ kg/hr} \] ### 2. Calculate the energy input per hour - Heating value: 40,000 kJ/kg - Total energy per hour: \[ 17.92 \text{ kg/hr} \times 40,000 \text{ kJ/kg} = 716,800 \text{ kJ/hr} \] ### 3. Convert power output to energy per hour - Power output: 55 kW - 1 kW = 1 kJ/sec - Energy per hour: \[ 55 \text{kW} \times 360 \text{ sec} = 198,000 \text{ kJ/hr} \] ### 4. Calculate efficiency \[ \text{Efficiency} = \frac{\text{Power output (energy per hour)}}{\text{Input energy per hour}} \times 100 \] \[ = \frac{198,000}{716,800} \times 100 \approx 27.6\% \] **Closest option: 26.5%** --- ### **Correct option:** **Option 4: 26.5%**