Instant cold packs consist of two bags; one containing water (c=4.184J/g * deg,
ID: 1043081 • Letter: I
Question
Instant cold packs consist of two bags; one containing water (c=4.184J/g * deg, DeltaH freezing= -6.01kJ/mol) and the other containing solid ammonium nitrate (m.m. = 80.043 g/mol). When the inner bag of water is broken by squeezing the package, the water dissolves the solid according to the following endothermic process: NH4NO3(s) -> NH4NO3(aq) DeltaH solution = 25.5 kJ/mol which quickly lowers the pack's temperature. If an instant cold pack contained 150.0mL of water (d=1.00 g/mL) at an initial temperature of 20.0 degrees celsius, how many grams of ammonium nitrate would need to dissolve to cool the water to 0 degrees celsius and freeze it? You may assume the cold pack is thermally isolated so no additional heat is exchanged with the environment.
Explanation / Answer
To start, we obtain the amount of heat that needs to be removed to freeze completely the 150 mL of water:
Qremoved: mCpdeltaT + ndeltaHfreezing
Where:
m -> is mass of water, which is 150 grams
Cp -> Heat capacity of water
n -> moles of water
deltaHfreezing -> heat of solidification or freezing
Qremoved: (150 grams * 4.184 J/g C * (0 - 20oC)) + (-6.01 kJ/mol * (150 grams * 1mol/18grams)
Qremoved: [(-12552 J) * (1 kJ / 1000 J)] - 50.08 kJ
Qremoved: -12.552 kJ - 50.08 kJ = -62.632 kJ
So we need to remove 62.632 kJ, which we'll remove with the heat of reaction from ammonium nitrate:
62.632 kJ = 25.5 kJ/mol * (x grams / 80.043 g/mol)
x = 196.60 grams of ammonium nitrate