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Physics

The efficiency of a device’s work is expressed as the ratio of the useful energy output of a
device to the total energy input in the system:

output of useful energy
Efficiency = x 100%
total energy input
Since power is the rate of energy transfer, it is also possible to calculate the efficiency using
the ratio of the useful power to the power consumed:
textbooks nis edu kz
output of useful power
Efficiency = x 100%
supplied power

ACT A kettle transfers about 120 000 joules of electrical energy to its heating coil in about
60 seconds. When heating 200 ml of water to 100 degrees Celsius, only about 70 000 J are
transferred as thermal energy (heat) to the water. What is the efficiency of the kettle?

Given SI Solution
Energy transferred to output of useful energy
Efficiency = x 100%
the water = 70 000 J J (joules) total energy consumed

Total energy input in 70 000 J
60 seconds = 120 000 J J (joules) Efficiency = 120 000 J x 100% = 58.3% → 60%
Efficiency = ?

Answer: Efficiency = 60%

ACT Let’s think about light bulbs again. In the case of a filament bulb, 120 J of electric
energy is transferred as 100 J of thermal energy (heat) and 20 J of light energy. For an
energy-saving bulb, 120 J of electric energy is spent on 10 J “lost” energy (heat), while 110 J
is light energy. Draw a labelled Sankey diagram to scale, and calculate the efficiency of each
light bulb.

ACT Discuss and analyze the following table. Why are the efficiencies of these devices
different? How do you think energy is transferred in these devices? Draw Sankey diagrams
to scale for two different devices.

Device Efficiency
fan heater 98%
electric engine 90%

diesel engine 30–40%

rocket engine 47%
petrol engine 20–30%

locomotive 7%–9%
steam engine 1%

Efficiencies of different devices

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