Chemistry Question

Question

You have a glass of water with a volume of 250 mL and a temperature of 12 degrees Celsius. You add a piece of dry ice, CO2(s), which has a mass of 25 g, to cool down his water. If dry ice has an enthalpy of sublimation of 25.2 kJ/mol, calculate the final temperature of the water after the dry ice has fully dissolved.

Answer 1

Heat lost by water = heat gained by dry ice

heat = q = mC∆T where m = mass, C = specific heat and ∆T = change in temperature.

Since dry ice is solid CO₂, when it sublimes, it becomes gaseous CO₂ and will not add significantly to the mass of the water (assumption). So we shall use the mass of H₂O as 250 ml x 1 g/ml = 250 g.

Heat gained by dry ice = mass x ∆Hsub = 25 g x 1 mol/44 g x 25.2 kJ/mol = 14.3 kJ = 14,300 J heat lost by water. This is a phase change for CO₂ and so there is no change in temperature (∆T)

Heat lost by water = (250 g)(4.184 J/gº)(∆T) = 14, 300 J

∆T = 14,300 J / (250 g)(4.184 J/gº) = 11.8º

Since the heat from the water went to sublime the solid CO₂, the temperature of the water will decrease by this amount. Final temperature of water = 12º - 11.8º = 0.2ºC