Question 2.2: A water heater contains 40 gallons of water. Water weighs 8.......
A water heater contains 40 gallons of water. Water weighs 8.34 lb per gallon. The present temperature of the water is 68ºF. The water must be raised to a temperature of 160ºF in one hour. How much power will be required to raise the water to the desired temperature?
First determine the weight of the water in the tank, because a BTU is the amount of heat required to raise the temperature of 1 pound of water 1 degree Fahrenheit:
40 gal \times 8.34 lb per gal = 333.6 lb
The second step is to determine how many degrees of temperature the water must be raised. This amount will be the difference between the present temperature and the desired temperature:
160ºF – 68ºF = 92ºF
The amount of heat required in BTU will be the product of the pounds of water and the desired increase in temperature:
333.6 lb \times 92ºF = 30,691.2 lb-degrees or BTU
1 Whr = 3.412 BTU/hr
Therefore,
\frac{30,691.2\ \mathrm{BTU}}{3.412\ \mathrm{BTU/hr}} = 8995.1 Whr