An experimental spacecraft consumes a special fuel at a rate of 2.10 × 102 L/min. The density of the fuel is 0.690 g/mL and the standard enthalpy of combustion of the fuel is –39.8 kJ/g. Calculate the maximum power (in units of kilowatts) that can be produced by this spacecraft. 1 kW = 1 kJ/s
You have 210 L/min Convert that to L/sec, use density to convert to g/sec and convert that to kJ/sec with enthalpy of combustion. Won't that answer be in kJ/sec = KW.
To calculate the maximum power that can be produced by the spacecraft, we need to determine the amount of fuel being consumed per second and then multiply it by the enthalpy of combustion.
First, let’s convert the fuel consumption rate from liters per minute to milliliters per second.
1 minute = 60 seconds
2.10 × 10^2 L/min = 2.10 × 10^2 L / 60 s = 3.50 × 10^0 L/s = 3.50 × 10^3 mL/s.
Now, since we know the density of the fuel is 0.690 g/mL, we can find the mass of the fuel being consumed per second.
Mass = volume × density
Mass = 3.50 × 10^3 mL/s × 0.690 g/mL = 2.415 × 10^3 g/s.
Finally, we can calculate the power using the formula:
Power = fuel consumption rate × enthalpy of combustion
Power = 2.415 × 10^3 g/s × (-39.8 kJ/g) = -96.137 × 10^3 kJ/s
Since we want the answer in kilowatts (kW), we can convert kJ/s to kW:
1 kW = 1 kJ/s
Power = (-96.137 × 10^3) kW = -96.137 kW
Therefore, the maximum power that can be produced by the spacecraft is approximately 96.137 kW.
To calculate the maximum power that can be produced by the spacecraft, we need to determine the rate at which energy is being consumed by the fuel.
First, let's calculate the mass flow rate of the fuel:
Mass flow rate = density × volume flow rate
Mass flow rate = 0.690 g/mL × 2.10 × 10^2 L/min
To convert the mass flow rate to grams per second (g/s), we need to convert the volume flow rate from L/min to mL/s:
1 L = 1000 mL
1 min = 60 s
Mass flow rate = 0.690 g/mL × (2.10 × 10^2 L/min × 1000 mL/1 L) × (1 min/60 s)
Mass flow rate = 0.690 g/mL × 2.10 × 10^2 × 1000 × 1/60 g/s
Now, let's calculate the power generated by the fuel:
Power = energy consumption rate / time
The energy consumption rate can be calculated using the standard enthalpy of combustion and the mass flow rate:
Energy consumption rate = enthalpy of combustion × mass flow rate
Energy consumption rate = -39.8 kJ/g × mass flow rate
Since 1 kW = 1 kJ/s, we can convert the energy consumption rate to kilowatts:
Power = (-39.8 kJ/g × mass flow rate) / 1000
Substituting the value of the mass flow rate and calculating the power, we get:
Power = (-39.8 kJ/g × 0.690 g/mL × 2.10 × 10^2 × 1000 × 1/60 g/s) / 1000
Simplifying the expression:
Power = (-39.8 kJ/g × 0.690 × 2.10 × 10^2 × 1/60) kW
Calculating the numerical value of the expression, we find:
Power ≈ -32.81 kW (rounded to the nearest hundredth)
Therefore, the maximum power that can be produced by this spacecraft is approximately -32.81 kW (negative value indicates that energy is being consumed instead of produced).