How much heat is needed to raise the temperature of 5.28 gal of water from 25 celsius degree to 88.0 celsius?

Question

(1 gal = 3.785)

On my calculation, I got 5.292 *10^-3, but I was wrong. I don't know How to solve this

Answer 1

Using the following equation:

Q = m*c*(Tf - Ti)

Where:

v = volume of water = 5.28 gal = 0.01999 m^3

m = mass of water = volume * density = 0.01999 m^3 * 1000kg/m^3 = 19.99 kg = 19986g

c = average heat capacity of water = 4.1813 J/(g*C)

Tf = 88C

Ti = 25C

Q = 19986g*4.1813 J/(g*C)*(88 - 25)C

Q = 5265kJ

Answer 2

Use the formula Q=mc(change in T)

Q = (19.99 L)  (4.186 J/g*C)  (63 C)

Q = 5271.7 kJ