Question

A 25 g glass tumbler contains 200 mL of water at 24°C. If two 15 g ice cubes each at a temperature of -3°C are dropped into the tumbler, what is the final temperature of the drink? Neglect thermal conduction between the tumbler and the room. 1°C

Answer 1

Consider that the system reaches a thermal equilibrium with a final temperature T.
Heat gained by ice cubes, Q1 = MiCiT
Where Mi is the mass of ice, Ci is the specific heat of ice and T is the change in temperature of ice
= 30 x 2.05 x (3 - 0)
= 184.5 J
Heat gained in the phase transformation from 0 ^oC ice to 0 ^oC water,
Q2 = MiLf
Where Lf is the latent heat of fusion of ice
= 30 x 333.55
= 10006.5 J
Heat gained when 0 ^oC water is warmed to T ^oC,
Q3 = Mi x Cw x (T - 0)
Where Cw is the specific heat of water
= 30 x 4.816 x T
= 144.48 T

Heat lost by glass tumbler
Q4 = Mg x Cg x (24 - T)
Where Mg and Cg are mass and specific heat of glass
= 25 x 0.84 x (24 - T)
= 21 (24 - T)
Heat lost by water,
Q5 = Mw x Cw x (24 - T)
Where Mw and Cw are mass and specific heat of water
= 200 x 4.186 x (24 - T)
= 837.2 (24 - T)

Total heat gained = Total heat lost
Q1 + Q2 + Q3 = Q4 + Q5
184.5 + 10006.5 + 144.48 T = 21 (24 - T) + 837.2 (24 - T)
10191 + 144.48 T = 858.2 (24 - T)
1002.68 T = 10405.8
T = 10.4 ^oC