Answer:
6.45 KWh
Explanation:
Data given:
amount of water = 58 gallon
Convert gallons to L
1 gallon = 3.78541 L
58 gallon = 58 x 3.78541 = 220 L
change in temperature = 25 °C
kilowatt-hours of energy = ?
Solution:
First we have to change volume of water into mass
following formula will use
m = d x v . . . . . . (1)
as we know
density of water = 1000 g/L
Put values in equation 1
m = 1000 g/L x 220 L
m = 220000
To calculate heat we use following formula
Q = m.c.ΔT . . . . . . . . . . (2)
Where
m= mass
c = specific heat of water
- specific heat of water = 4.186 J/g °C
ΔT = change in temperature
Put values in equation 1
Q = 220,000 g x 4.186 J/g °C x 25 °C
Q = 23023000 J
As we need this value in kilowatt-hours of energy so we convert J to KWh
1 J = 2.8 x 10⁻⁷ KWh
23023000 J = X KWh
Do cross multiplication
KWh = 23023000 J x 2.8 x 10⁻⁷ KWh / 1 J
KWh = 6.45 KWh
6.45 KWh (kilowatt-hours) of energy are needed to heat the water in the water heater.
