Question

a water heater contains 58 Gal of water. How many kilowatt-hours of energy are necessary to heat the water in the water heater by 25 ∘C?

Answer 1

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.