Efficiency, Work and power
1) A light bulb takes in 30J of energy per second. It transfers 3J as useful light
energy and 27J as heat energy. Calculate the efficiency.
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
𝑢𝑠𝑒𝑓𝑢𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑢𝑡
𝑇𝑜𝑡𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑖𝑛
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
3
= 0.1
30
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
𝑢𝑠𝑒𝑓𝑢𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑢𝑡
𝑇𝑜𝑡𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑖𝑛
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
1500 + 1000
= 0.5
5000
5) Calculate the efficiency of a light bulb that gives of 40J of light from 200J of
electrical energy.
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
Efficiency as a percentage = 0.1 x 100 = 10%
2) A kettle takes in 2000J of energy per second. It transfers 1500J as useful
heat energy and 500J is wasted as sound energy. Calculate the efficiency of
the kettle.
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
𝑢𝑠𝑒𝑓𝑢𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑢𝑡
𝑇𝑜𝑡𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑖𝑛
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
1500
= 0.75
2000
Efficiency as a percentage = 0.75 x 100 = 75%
3) Calculate the efficiency of a hair dryer which takes in 3000J of energy per
second and transfers 600J as useful heat energy. Express your answer as a
decimal and not as a percentage.
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
𝑢𝑠𝑒𝑓𝑢𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑢𝑡
𝑇𝑜𝑡𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑖𝑛
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
600
= 0.2
3000
4) Calculate the efficiency of a TV which takes in 5000J of energy per second
and transfers 1000J as useful light energy and 1500J as useful sound energy.
The remaining 2500J is wasted as heat energy. Express your answer as a
decimal and not a percentage.
𝑢𝑠𝑒𝑓𝑢𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑢𝑡
𝑇𝑜𝑡𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑖𝑛
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
40
= 0.2
200
Efficiency as a percentage = 0.2 x 100 = 20%
6) Calculate the efficiency of a radio that gives of 30J of sound from 360J of
electrical energy.
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
𝑢𝑠𝑒𝑓𝑢𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑢𝑡
𝑇𝑜𝑡𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑖𝑛
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
30
= 0.083
360
Efficiency as a percentage = 0.083 x 100 = 8.3%
7) Calculate the efficiency of a runner who produces of 500J of kinetic energy
from 2500J of chemical energy.
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
𝑢𝑠𝑒𝑓𝑢𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑢𝑡
𝑇𝑜𝑡𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑖𝑛
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
500
= 0.2
2500
Efficiency as a percentage = 0.2 x 100 = 20%
8) Calculate the percentage efficiency of a student who takes 2 hours to do 30
minutes of homework.
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
𝑢𝑠𝑒𝑓𝑢𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑢𝑡
𝑇𝑜𝑡𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑖𝑛
12) What is the useful work done by a runner of efficiency 0.30 when supplied
with 200J of energy?
30
= 0.25
120
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
Efficiency as a percentage = 0.25 x 100 = 25%
0.3 =
9) What is the useful light output from a bulb of efficiency 0.25 when supplied
with 600J of energy?
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
𝑢𝑠𝑒𝑓𝑢𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑢𝑡
𝑇𝑜𝑡𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑖𝑛
0.25 =
13) How long does it take a student of efficiency 40% to do a 30 minute
homework?
?
600
0.6 =
?
1200
𝑜𝑢𝑡𝑝𝑢𝑡 = 1200 𝑥 0.6 = 720𝐽
a.
30
𝑡𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒
𝑡𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒 =
30
= 75 𝑚𝑖𝑛𝑠
0.4
How much electrical energy does this bulb use in one second?
40W
b.
Calculate the efficiency of the bulb if it gives out 5J of light in one
second.
𝑢𝑠𝑒𝑓𝑢𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑢𝑡
𝑇𝑜𝑡𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑖𝑛
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
𝑢𝑠𝑒𝑓𝑢𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑢𝑡
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
𝑇𝑜𝑡𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑖𝑛
𝑜𝑢𝑡𝑝𝑢𝑡 = 20 𝑥 0.5 = 10𝐽
40% = 0.4 =
14) A light bulb is rated as 40W
11) What is the useful sound output from a radio of efficiency 50% when
supplied with 20J of energy?
?
50% = 0.5 =
20
𝑢𝑠𝑒𝑓𝑢𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑢𝑡
𝑇𝑜𝑡𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑖𝑛
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
10) What is the useful work output from a motor of efficiency 0.60 when
supplied with 1200J of energy?
𝑢𝑠𝑒𝑓𝑢𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑢𝑡
𝑇𝑜𝑡𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑖𝑛
?
200
𝑜𝑢𝑡𝑝𝑢𝑡 = 200 𝑥 0.3 = 60𝐽
𝑜𝑢𝑡𝑝𝑢𝑡 = 600 𝑥 0.25 = 150𝐽
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
𝑢𝑠𝑒𝑓𝑢𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑢𝑡
𝑇𝑜𝑡𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑖𝑛
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
5
remember 40W = 40 J/s
40
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 = 0.125
c.
What is the percentage efficiency of the bulb?
0.125 x 100 = 12.5%
d.
What is the useful power output of the light bulb?
a.
The runner’s kinetic energy
5W
e.
𝐾𝐸 =
How much light energy should the bulb give out in 3 minutes?
5W x (3 x 60) = 900J
f.
𝐾𝐸 =
How much heat energy should the bulb give out in 3 minutes?
Heat energy / sec = 40W – 5W = 35W
a.
What is the weight of the mass?
W = mg = 4 x 10 = 40N
b.
How much work does the motor do?
Work = force x distance = 40 x 3 = 120J
c.
What is the power of the motor?
Power = work / time = 120 / 5 = 24W
d.
What is the percentage efficiency of the motor?
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
𝑢𝑠𝑒𝑓𝑢𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑢𝑡
𝑇𝑜𝑡𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑖𝑛
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
120
= 0.33
360
Efficiency as a percentage = 0.33 x 100 = 33.3%
e.
What is the power supplied to the motor?
Power = energy supplied / time = 360 / 5 = 72W
16) A runner of mass 120kg maintains a speed of 5 m/s while using chemical
energy from food at a rate of 4500J per second. Calculate:
120 𝑥 52
2
KE = 1500J
b.
35W x 180 = 6300W (6.3kW)
15) An electric motor pulls a mass of 4kg up by 3 metres in 5 seconds when
supplied with 360J of electrical energy.
𝑚𝑣 2
2
The input power to the runner from food
Power = 4500 J/s = 4500W
c.
The efficiency of the runner
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
𝑢𝑠𝑒𝑓𝑢𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑢𝑡
𝑇𝑜𝑡𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑖𝑛
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
1500
= 0.33
4500