2) (a) Two 84.5 g ice cubes are dropped into 30 g of water in a glass.  If the water is initially at a temperature of 5⁰ C and if the ice comes directly from  the freezer at -30⁰ C, what will be temperature of the drink when the ice and the water reach thermal equilibrium ?  How much ice and how much water are present at thermal equilibrium ?  Ignore the heat capacity of the glass and heat transferred to and from the environment.  For full credit carefully explain each step of your analysis. (13)

T_H = 5⁰ C ; T_C = -30⁰ C ; m_w = 30 g ; m_i = 170 g ; T₀ = 0⁰ C

Heat lost by water = Heat gained by ice

Heat provided by water in cooling to 0⁰ C = m_wc_w(T_H –T₀) = 0.03 x 4190 x 5 = 628.5 J

Heat provided by water in freezing = m_wL_F = 0.03 x 334 x 10³ = 10020 J

Total heat provided by water = 10020 + 628.5 = 10648.5 J

Heat to raise ice from -30⁰ C to 0⁰ C = m_ic_i(T₀ – T_c) = 0.169 x 2100 x 30 = 10647 J

Therefore, heat provided by water in cooling and freezing is (almost) exactly equal to the heat needed to bring the ice to 0⁰ C.

Equilibrium temperature is 0⁰C. No water and 0.199 kg ice (2 x 0.0845 + 0.03).

(b) Suppose that only one ice cube had been used in part (a), what would be the final temperature of the drink ?   How much ice and how much water are present at thermal equilibrium ?  Ignore the heat capacity of the glass and heat transferred to and from the environment. (12)

With a single ice cube, the only change from above is the heat needed to raise the ice to 0⁰ C from -30⁰ C, which is now one half the value above = 0.5 x 10647 = 5323.5 J

Cooling the water to 0⁰ C from 5⁰ C provides only 628.5 J of this heat. Only enough water to provide (5323.5 – 628.5) 4695 J of heat by freezing is needed to bring the ice to 0⁰ C.

Mass of water that freezes = 4695/L_F = 4695/(334 x 10³) = 0.014 kg (14 g)

Thus, equilibrium temperature is 0⁰ C at which time there is 14 + 84.5 = 98.5 g of ice present and 30-14 = 16 g of water.