Question

Two 20.0-g ice cubes at –15.0 °C are placed into 265 g of water at 25.0 °C. Assuming no energy is transferred to or from the surroundings, calculate the final temperature, Tf, of the water after all the ice melts.

Answer 1

Two ice cubes weighs 40 g at –15.0 °C are placed into 265 g of water at 25.0 °C.

The amount of heat absorbed by two ice cubes, Q₁ = (2x20 g) x 15 ^oC x 2.027 J/g = 1216.2 J
Latent heat of fusion for two ice cubes,Q₂ = 40 g x 334 J/g =13360 J
Heat absorbed by melting of two ice cubes,Q₃ = 40 g x (T_f - 0 ^oC) x 4.184 J/g^oC (here T_f is the final temperature)

Q₃ = 167.36T_f
Heat released by water, Q₄ = 265 x (25 - T_f) x 4.184 J/g^oC = 27719 - 1108.76T_f
Q₄ = Q₁+Q₂+Q₃
27719 - 1108.76T_f = 1216.2 J + 13360 J + 167.36T_f
13142.8 J = 1276.12T_f

T_f = 10.29°C