specific heat--chemistry problem?

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 endothermic reaction:

NH4NO3 (s) + H2O (l) -> NH4NO3 (aq)
DeltaH= +25.7 kj

What is the final temperature in a squeezed cold pack that contains 50 g of NH4NO3 dissolved in 125 mL of water? Assume the specific heat is 4.18 j/gc for the solution, an initial temperature of 25C and no heat transfer between the cold pack and the environment?


I AM SO LOST CAN SOMEONE HELP ME SET UP AND SOLVE THIS CHEMISTRY PROBLEM. IM GOING CRAZY!

Answer 1

delta H = sp.heat X mass X delta T

delta H you know (1000j = 1 kj)
sp. heat you know (The assumption is that the solution has the specific heat of water, 1 cal/gC = 4.18 J/gC)
mass = 125 g water + 50 g ammonium nitrate
You know the mass of water b/c density of water = 1 g/mL
delta T is the change from T(initial) (25C) to some final value.

So you need to rearrange the equation to
delta T = delta H/(mass X sp.ht)

And finally, delta T = T(final) - T(initial), so T(final) = delta T + T(initial).

That should do it...

Answer 2

First find the moles of NH4NO3. The molar mass of NaNO3 equals 80 g/mol:

n = m/Mr, n = 50/80, n = 0.625 mol

When 1 mol of NH4NO3 is dissolved 25.7 kJ or 25700 J of energy is absorbed, so when 0.625 mol of NH4NO3 are dissolved in the water:

q = 0.625*25700, q = 16,062.5 J of energy will absorbed.

Find the mass of the solution:

m = 50 + 125 = 175 g (because the density of the water is 1 g/ml so 125 mL of water are 125 g of water)

Now: q = m*c*ΔT, so ΔT = q/m*c, ΔT = 16062.5/(175*4.18)

ΔT = 21.9 deg. or ΔT = 22 deg. approx.

So the final temperature will be:

Tf = Ti - ΔT, Tf = 25 - 22, Tf = 3 oC