How much work can a 22 kW car engine do in 60 seconds if
It is 30% efficient
It is 100% efficient
To calculate the work done by the car engine, we can use the formula:
Work = Power x Time
1. For a 30% efficient engine:
Efficiency = 30% = 0.30
Power = 22 kW
Time = 60 seconds
Work = Power x Time
Work = 22 kW x 60 s = 1320 kWs
Since the engine is only 30% efficient, the work done is 1320 kWs x 0.30 = 396 kWs.
2. For a 100% efficient engine:
Efficiency = 100% = 1.00
Power = 22 kW
Time = 60 seconds
Work = Power x Time
Work = 22 kW x 60 s = 1320 kWs
Since the engine is 100% efficient, the work done is 1320 kWs.
To find out how much work a car engine can do in 60 seconds, we'll calculate it separately for two scenarios: when the engine is 30% efficient and when it is 100% efficient.
1. When the engine is 30% efficient:
Efficiency is defined as the ratio of useful work output to the total energy input. So if the engine is 30% efficient, it means that only 30% of the energy input is converted into useful work. The remaining 70% is lost as waste heat or other forms of energy.
First, let's calculate the energy input to the engine in 60 seconds:
Energy input (in joules) = Power (in watts) × Time (in seconds)
Power = 22 kW = 22,000 watts
Time = 60 seconds
Energy input = 22,000 watts × 60 seconds = 1,320,000 joules
Now, to find the useful work output, which is 30% of the energy input:
Useful work output = Efficiency × Energy input
Efficiency = 30% = 0.3
Useful work output = 0.3 × 1,320,000 joules = 396,000 joules
Therefore, when the engine is 30% efficient, it can do 396,000 joules of work in 60 seconds.
2. When the engine is 100% efficient:
In this scenario, the engine is 100% efficient, meaning all the energy input is converted into useful work, and none is wasted.
Using the same energy input in 60 seconds as calculated above:
Energy input = 1,320,000 joules
Since the engine is 100% efficient, the useful work output is equal to the energy input:
Useful work output = Energy input = 1,320,000 joules
Therefore, when the engine is 100% efficient, it can do 1,320,000 joules of work in 60 seconds.