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Difficulty: MediumWork, Energy, and Power

An industrial pump performs 72,000 J72,000\text{ J} of work to circulate coolant through a cooling system in 2 minutes2\text{ minutes}. What is the average power output of the pump?

  1. A
    36,000 W36,000\text{ W}
  2. 600 W600\text{ W}Answer
  3. C
    144,000 W144,000\text{ W}
  4. D
    72,000 W72,000\text{ W}

Answer

600 W600\text{ W}
Power is defined as the rate at which work is done (P=WtP = \frac{W}{t}). The standard unit of power, the watt (W\text{W}), equals one joule per second (J/s\text{J/s}). Converting 2 minutes2\text{ minutes} gives 120 seconds120\text{ seconds}. Dividing 72,000 J72,000\text{ J} by 120 s120\text{ s} gives an average power output of 600 W600\text{ W}.

Step-by-Step Solution

1
Convert the elapsed time from minutes into the standard SI unit of seconds
2 minutes×60 seconds/minute=120 seconds2\text{ minutes} \times 60\text{ seconds/minute} = 120\text{ seconds}
The watt is defined as one joule per second, requiring time to be in seconds.
2
Apply the power formula P=WtP = \frac{W}{t} by dividing total work by time
P=72,000 J120 s=600 WP = \frac{72,000\text{ J}}{120\text{ s}} = 600\text{ W}
Power measures the rate at which work is performed over time.

Key Concept

Power as the Rate of Doing Work (P=WtP = \frac{W}{t})
Estimated Time:1m 0s
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Topics

Work, Energy, and Power

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