Question

Consider a glass of 189 mL of water at 27°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

Answer – We are given the , volume of water = 189 mL , ti = 27^oC,

Initial temp of ice ti = -15^oC, final temp = 10^oC

Mass of water = 189 mL * 1.00 g/mL

                       = 189 g

Heat lost by water = heat gain by ice

Heat of water , q = m * C*∆t

                            = 189 g * 4.184 J/g^oC * (10-27)^oC

                            = - 13301 J

So, heat of ice absorbed = 13301 J. Assume no heat lost

So, heat of the ice = heat needed -15 ^oC to 0.0^oC + heat of fusion + heat from the 0.0^oC to 10^oC

     q ice = q1+q2+q3

             = m* 2.03 g/^oC*(0.0-(15) + m* 334 J/g + m *4.184J/g^oC * (10.0-0)

13301 J = 30.45 m + 334m + 41.84 m

            = 406.29 m

So, m = 13301 / 406.29

           = 32.73 g

So, mass of ice = 32.73 g