Question

An experimental spacecraft consumes a special fuel at a rate of 372 L/min. The density of the fuel is 0.730 g/mL and the standard enthalpy of combustion of the fuel is − 26.5kJ/g. Calculate the maximum power (in units of kilowatts) that can be produced by this spacecraft. 1kW = 1kJ/s?

Answer 1

Explanation:

First, we will calculate fuel consumption is as follows.

         $372 L/min \times 1000 ml/L \times 0.730 g/ml \times \frac{1}{60} min/s$

           = 4526 g/s

Now, we will calculate the power as follows.

        Power = Fuel consumption rate × -enthalpy of combustion

                    = $4526 g/s \times -26.5 kJ/g$

                    = $1.19 \times 10^{5}$ kW

Thus, we can conclude that maximum power (in units of kilowatts) that can be produced by this spacecraft is $1.19 \times 10^{5}$ kW.