A60-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.

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

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form