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 10^{6} 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 10^{6} J is given by

C = ( 25 cents / 3600000 J ) × 10^{6} = 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 10^{6} 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,

10^{6} Joules = 10^{6} × (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 × 10^{6}

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 × 10^{6} = m × (100 - 40) × 4180 J/kg°C

Solve for m

m = 72 × 10^{6} / (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 -

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