Question

An experimental spacecraft consumes a special fuel at a rate of 353 L/min . The density of the fuel is 0.700 g/mL and the standard enthalpy of combustion of the fuel is − 57.9 kJ/g . Calculate the maximum power (in units of kilowatts) that can be produced by this spacecraft. 1 kW = 1 kJ/s

Answer 1

Step-by-step 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,
`353 * (1000 ml)/(60 sec)`

= 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.

`5883.33 ml * 0.7 g/ml`

= 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.