Question

Two 20.0‑g ice cubes at −17.0 °C−17.0 °C are placed into 215 g215 g of water at 25.0 °C. Assuming no energy is transferred to or from the surroundings, calculate the final temperature of the water after all the ice melts.

Answer 1

mass of ice = 2 x 20 = 40 g

moles of ice = 40 / 18.02 = 2.22 mol

Q1 = - m Cp dT

     = - 2.22 x 37.7 x 17

     = - 1422.8 J

Q2 = 2.22 x 6.01 x 10^3

      = -13342.2 J

Q3 = 2.22 x 75.3 x 25

      = -4179.15 J

total heat = Q1 + Q2 + Q3

                = -18944 J

Q = m Cp dT

moles of water = 40 + 215 / 18.02 = 14.15

-18944 = 14.15 x 75.3 x (Tf - 25)

Tf = 7.22 oC

final temperature = 7.22 ^oC