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A 60-gallon water heater is initially filled with water at 70°F. Determine how much energy, in Btu, needs to be transferred to the water to raise its temperature to 100°F.Given: The water properties at an average water temperature of 85ºFrho = 62.17 lbm/ft3 and cp = 0.999 Btu/lbm-R.

Respuesta :

Answer:

Q = 14948.73 Btu

Explanation:

data given:

volume of water = 60 gallon = 60*0.133681 = 8.02083 ft3

temperature t2 = 100°F

temperature t1 = 70°F

At 85°F density of water is 62.17 lb m/ft3

cp = 0.999 Btu/lbm-R

Change in temperature = 100 - 70 = 30°F = 30R

From the above equation, total energy needed is

Q = 62.17 *8.02083*0.999*30

Q = 14948.73 Btu

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