Water heater contains 51 gal of water. part a how many kilowatt-hours of energy are necessary to heat the water in the water heater by 25 ∘c?

Answer is: 550,021 kWh of energy is needed to heat the water

V (water) = 51 gal = 51·3,78 = 189,3 L.

ΔT (water) = 25°C.

d (water) = 1000 g/L.

m (water) = V (water) · d (water)

m (water) = 189,3 L · 1000 g/L

m (water) = 189300 g.

Q = m (water) · ΔT (water) · C (water)

Q = 189300 g · 25°C · 4,184 J/°C·g

Q = 19800780 J = 19800,78 kJ : 3600 = 550,021 kWh.

5.6 kilo-watt hours

The specific heat of water is 4.1796 J / (K*cm^3) which means that in order to raise the temperature of 1 cubic centimeter of water 1 Kelvin, it takes 4.1796 Joules of energy. So let's start by converting 51 gallons into cubic centimeters:

51 gal * 3.78541 L/gal * 1000 cm^3/L = 193055.91 cm^3

Since the size of 1 Kelvin is the same as the size of 1 degree C, we don't need to worry about converting the temperature. 25 degree C increase is the same as a 25 K increase. So let's calculate how many joules we need.

193055.91 cm^3 * 25 K * 4.1796 J / (K*cm^3) = 20172412.04 J

20172412.04 J = 20172412.04 kg*m^2/s^2

20172412.04 kg*m^2/s^2 / 3600 s/h = 5603.447788 watt hours. 5603 / 1000 = 5.6 kilowatt-hours.