Question

Some sports ‘cold packs’ use the dissolution of ammonium nitrate as the cooling source.Here the solid ammonium nitrate package is broken, and cools as it dissolves. Calculate the final temperature of a cold pack which contains 100 g of NH4NO3 and 400 g water at 25 °C.Assume the specific heat of the ammonium nitrate solution is 4.0 J/g-C. NH4NO3 80.0 g/mol. NH4NO3(s) → NH4NO3(aq) DrHo = +25.7 kJ/mol

Answer 1

Amount of heat absorbed by the reaction = amount of heat released from the water

Given enthalpy of reaction = ΔH° = 25.7 kJ/mol

Amount of heat liberated from 100 g NH₄NO₃ =

=32.125 kJ

Amount of heat liberated from 100 g NH₄NO₃ = = 32.125 kJ

This heat is released by the water

so

Mass of ammonium nitrate solution = m = 100 + 400 = 500 g

specific heat of the solution = c = 4.0 J/g.^oC

calculating change in temperature =

divide both sides with 500 x 4.0

Given initial temperature = T_i = 25 ^oC

Final temperature = T_f = ΔT + T_i = 25 -16 = 9 ^oC

Final temperature = T_f = 9 ^oC