Question

Consider a glass of 167 mL of water at 29°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.

Answer 1

Heat gained by ice = heat lost by water

                   mcdt = m'c'dt'

Where

m= mass of ice added = ?

c = specific heat capacity of ice = 2.09 J/goC

dt = change in temperature of ice = final - initial = 10^oC - (-15^oC) = 25^oC

m' = mass of water = volume x density = 167 mL x 1.0 g/mL = 167 g

c = specific heat capacity of water = 4.186 J/g^oC

dt = change in temperature of ice = initial - final = 29^oC - 10^oC = 19^oC

Plug the values we get

m x 2.09 x 25 = 167 x 4.186x19

                 m = 254.2 g

Therefore the mass of ice added is 254.2 g