Question

A hot tub is filled with 410 gal water. How many kilocalories are needed to heat the water from 67 ∘F to 103 ∘F? (Assume that 1cal=4.184J, the specific heat of water is 4.184 J/g∘C.)

Answer 1

volume of water = 410 gal = 410 x3.78541

                        = 1552 L

                        = 1552 x 10^3 mL

                       = 1.55 x 10^6 mL

density of water = 1 g / mL

mass of water = 1 x 1.55 x 10^6

                     = 1.55 x 10^6 g

temperature = t1 = 67 F = 19.4 oC

temperature = t2 = 103 ∘F = 39.4 oC

dT = t2-t1 = 39.4 - 19.4 = 20 oC

Q = m Cp dT

Q = (1.55 x 10^6) (1) (20)

Q = 3.10 x 10^7 cal

Q = 3.10 x 10^4 kcal