Explanation:
The given data is as follows.
Space craft fuel rate = 353 L/min
As 1 liter equals 1000 ml and 1 min equals 60 seconds.
So, [tex]353 \times \frac{1000 ml}{60 sec}[/tex]
= 5883.33 ml/sec
It is also given that density of the fuel is 0.7 g/ml and standard enthalpy of combustion of fuel is -57.9 kJ/g.
Fuel rate per second is 5883.33 ml.
[tex]5883.33 ml \times 0.7 g/ml[/tex]
= 4118.33 g
Hence, calculate the maximum power as follows.
Power = Fuel consumption rate × (-enthalpy of combustion)
= 4118.33 g/s \times 57.9 kJ/g
= 238451.36 kJ/s
or, = 238451.36 kW
Thus, we can conclude that maximum power produced by given spacecraft is 238451.36 kW.
