Question

Consider a glass of 184 mL of water at 28°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

Volume of water = 184 mL

density of water = 1.0 g/mL

Hence, mass of water, m = 184 g

Given that temeperature of the water to be lowered from 28^∘C to 10^∘C.

temeperature of ice to be increased from -15^∘C to 10^∘C.

Heat lost / gained Q = mc dT

Heat lost by water = Heat gained by ice

mc (t1-t) = mc (t-t2)

(184 g) x (4.184 J/g/^oC) x [ (28^oC-10^oC)] = m x 2.03 J/g/^oC x [(10^∘C - (-15^∘C)]

m = 273 g

mass of ice = 273 g

Therefore, 273 grams of ice at -15°C must be added to cool the water to 10°C.