A60-gallon water heater is initially filled with water at 66°f. determine how much energy, in btu, needs to be transferred to the water to raise its temperature to 104°f. given: the water properties at an average water temperature of 85ºf rho = 62.17 lbm/ft3 and cp = 0.999 btu/lbm-r.

18930 Btu

Explanation:

The amount of heat needed to raise the temperature of a body of water is given by:

Q = m * cp * (tfin - ti)

The mass is the volume multiplied by the density

m = V * ρ

So

Q = V * ρ * cp * (tfin - ti)

The volume of water is 60 gallon, which is equivalent to 8.021 ft^3

Now:

Q = 8.021 ft^3 * 62.17 lbm/(ft^3) * 0.999 Btu/(lbm * R)* (104 R - 66 R) = 18930 Btu

Since we have a temperature difference we can convert Fahrenheit to Rankine

i think it would be 5

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