Can anyone help me solve this problem? Thanks. I coach U10 soccer and go through

Question

Can anyone help me solve this problem? Thanks.

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_4NO_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 solution for ammonium nitrate in water.

Explanation / Answer

Density of water = 1g/mL

Volume of water taken = 300 mL

So, mass of water taken = 300*1 = 300 g

Total heat released by water in this process =

Q = mCp(water)dT1 + mL + mCp(ice)dT2

Cp(water) = 4.18 J/(g.K)

Cp(ice) = 2.1 J/(g.K)

L = 334 J/g

m = 300 g

dT1 = (30-0) = 30 K

dT2 = (0-(-2)) = 2 K

Putting values we get:

Q = 300*4.18*30 + 300*334 + 300*2.1*2 = 139080 J

Specific heat of solvation for this solute in water = Q/m = 139080/300 = 463.6 J/g

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