Question

Consider a glass of 166 mL of water at 30°C. Calculate the mass of ice at -15°C that must be added to cool the water to 10°C after thermal equilibrium is achieved. To find the mass of water use the density of water = 1.0 g/mL.
g

Answer 1

The amount heat required for the conversion of water at 30 oC to water at 10 oC is , Q = mcdt

Where

m = mass of water = volume x density = 166 mL x 1.0 g/mL = 166 g

c = specific heat capacity of water = 4.186 J/g degree C

dt = change in temperature = 30 - 10 = 20 oC

So Q = 166 x 4.186 x 20

= 13.897x10³ J

Heat gained by ice = heat lost by water

heat change for conversion of ice at -15 ^oC to ice at 0 ^oC + heat change for onversion of ice at 0^oC to water at 0^oC + heat change for conversion of water at 0^oC to water at 10 ^oC = heat lost by water = 13.897x10³ J

m'c'dt' + m'L + m'cd''t = 13.897x10³ J

m'(c'dt' + L + cdt'') = 13.897x10³ J

Where

m' = mass of ice required = ?

c' = Specific heat of ice= 2.09 J/g degree C

L = Heat of fusion of ice = 334.9 J/g

dt' = 0-(-15) = 15 oC

dt'' = 10-0 = 10 oC

Plug the values we get

m' = 34.0 g

So the mass of ice added to cool the water is 34.0 g