An online conversion calculator of the unit of energy Watt-hour (Wh) and Joules (J) is presented. Examples and problems involving conversion of Watt-hour and Joules are also included.
The Watt-hour (Wh) and Joules (J) are units of energy with conversion rate as follows:
1 Watt-hour (Wh) = 3600 Joules (J)
1 Joule J) = (1 / 3600) Watt-hour (Wh)
Example 1 - Convert 15.6 Watt-hour (Wh) to Joules (J).
Given that 1 Wh = 3600 (J)
15.6 Wh = 15.6 × 3600 (J) = 56160 J
Example 2 - Convert 5660 Joules to Watt-hour (Wh) and round to the nearest tenth of a Wh
Given that 1 Joule J) = (1 / 3600) Watt-hour (Wh),
5660 J = 5660 * (1 / 3600) Wh = 1.57222222222 Wh
Round to the nearest tenth of a Wh
5660 J = 1.6 Wh
Enter the number of Watt-hours (Wh) or the number of Joules (J) and convert.
Problem 1
The cost of 1000 Wh of electricty is 25 cents. How much does 106 Joules of eletcricty cost? Round answer to the nearest cent.
Solution
Given that 1 Wh = 3600 J
1000 Wh = 1000 × 3600 J = 3600000 J
The cost of 1000 Wh is also the cost of 3600000 J and it is equal to 25 cents. The cost C of 106 J is given by
C = ( 25 cents / 3600000 J ) × 106 = 6.94444444444 cents
Round to the nearest cent.
C = 7 cents.
Problem 2
It takes 4180 Joules to raise the temperature of 1 Kg of water by 1 degree celsius. How many Watt-hours are needed to raise the temperature of 1 gram of water by 1 degree celsius? Round to the nearest tenth.
Solution
Given that 1 J = (1 / 3600) Wh
4180 J = 4180 × (1 / 3600) Wh = 1.16111111111 Wh
Round to the nearest tenth
1.2 Wh are needed to raise the temperature of 1 Kg of water by 1 degree celsius
Problem 3
One cubic foot of natural gas can produce about 106 Joules of energy. How many Watt-hours may be produced by 1 cubic foot of natural gas? Round to the nearest unit.
Solution
Given that 1 J = (1 / 3600) Wh,
106 Joules = 106 × (1 / 3600) Wh = 277.777777778 Wh
Rund to the nearest unit
278 Wh of energy may be produced by 1 cubic foot of natural gas.
Problem 4
The specific heat capacity of water is 4180 J/kg°C, which means you need 4180 J to raise the temperature of 1 Kg of water by 1 °C.
A domestic solar panel system produces on average 20000 Wh of eletcricity per day. How many Kg of water are heated from 40°C to 100°C per day by this system? Round to the nearest Kg.
Hint: the formula that gives the energy E needed to raise m kg of water from temperature T1 to T2 is given by: E = m × (4180 J/kg°C) × (T2 - T1).
Solution
Since the heat capacity of water is given in J/kg°C, we first need to convert the energy produced per day from Wh to J. Hence given that 1 Wh = 3600 J
20000 Wh = 20000 × 3600 J = 72 × 106
Let m be the number of kg of water heated from 40°C to 100°C. Using the formula for the energy E given above, we write
72 × 106 = m × (100 - 40) × 4180 J/kg°C
Solve for m
m = 72 × 106 / (60 × 4180)
Evaluate m and round to the nearest Kg
m = 287 Kg of water are heated from 40°C to 100°C per day by this solar system.
Example 2 -
