I coach U10 soccer and go through an insane amount of instant ice packs. I have

Question

I coach U10 soccer and go through an insane amount of instant ice packs. I have considered investing in magic spray, but the soccer club will pay for the ice pack and not the spray and I am super cheap. So, clearly, I was curious how they work. Turns out that most of them are a small package of ammonium nitrate (NH_4 NO_3) in water. When the inside bag is broken, the solute dissolves in the water. This dissolution requires energy and the water cools and freezes. I have done many experiments to find the best packs, and the market leader uses 300 g of the solute and 300 mL of water. It reaches a final temperature of-2 C. If I assume the dissolution process is so fast that it can be modelled as adiabatic, calculate the specific heat of solvation for ammonium nitrate in water.

Explanation / Answer

Ans. Let the initial temperature of ice pack = 250C        - be it at room temperature

                        Final temperature of ice pack = -20C        - given

                        Mass of water = 300 g                                             - given

                        Mass of NH4NO3     = 300 g                                  - given

                        Heat capacity of water = 4.18 J/ 0C

When NH4NO3 mixed with water, the temperature of water drops from 298 K to 271 K. So,

Change in temperature, dT = (initial - final) temperature

                                                = 250C – (-20C)

                                                = 270C

It’s assumed that the resultant solution remains in liquid state. Also, the heat capacity of the solution is the same as water.

While cooling down by 270C, water must lose heat.

Total heat lost by water, q = m x s x dT             - equation 1   

Where,

m = mass in gram,

c = specific heat of water = 4.18 J g-10C-1

dT = change in temperature = (final – initial) temperature

            or, q = 300 g x 4.18 J g-10C-1 x 270C

                        = 33858 J

                        = 33.858 kJ

Thus, while cooling water loses 33.858 kJ heat.

Now,

Heat lost by water is gained by ammonium nitrate for solvation.

Moles of NH4NO3 in 300 g sample = Mass/ molar mass

                                                            = 300 g / (80.0 g mol-1)

                                                            = 3.75 mol

Specific heat of solvation = heat gained / mass of NH4NO3

                                    = 33858 J / 300 g

                                    = 112.86 J = 0.11286 J/g

Specific heat (molar) of solvation = heat gained / moles of NH4NO3

                                    = 33858 J / 3.75 mol

                                    = 9028.8 J/mol = 9.02 J/mol

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