Instant cold packs used to treat athletic injuries contain solid NH 4 NO 3 and a
ID: 981176 • Letter: I
Question
Instant cold packs used to treat athletic injuries contain solid NH4NO3 and a pouch of water. When the pack is squeezed, the pouch breaks and the solid dissolves, lowering the temperature because of the following endothermic reaction.
What is the final temperature in a squeezed cold pack that contains 40.0 g of NH4NO3 dissolved in 125 mL of water? Assume a specific heat of 4.18 J/(g·°C) for the solution, an initial temperature of 25.0°C, and no heat transfer between the cold pack and the environment. To find the mass of water use the density of water = 1.0 g/mL. Hint: The process takes place at constant pressure.
Explanation / Answer
The equation can be rewritten as
NH4NO3(s) + H2O (l) + 25.7 kJ NH4NO3(aq)
For one mole of NH4NO3, the amount of heat absorbed = 25.7 kJ (for 80 grams ie, 80 gram = 1mole).
For 40.0 grams of NH4NO3, the amount of heat absorbed can be calculated as follows:
40*25.7/80 = 12.85 kJ
q = sp. heat * mass * Delta T
q = -12.85*1000 J = -12850 J
125 mL of water = 125 g of water
Thus
-12850 = 125*4.18* delta T
delta T = -24.58
T (final) - T (initial) = -24.58
T(final) = T (initial) - 24.58
= 25 - 24.58
= 0.42 oC
Thus