Question

A cold pack manufacturer has decided to change the chemical in their cold pack from ammonium nitrate to ammonium chloride, a safer chemical. The original cold pack formulation used 100.0 g of room temperature (27.0 oC) water and 35.0 g of ammonium nitrate. Assume that the solution has the same density and specific heat capacity of water and that any heat lost to the calorimeter is negligible. The molar enthalpy of solution of ammonium nitrate is 25.7 kJ/mole and for ammonium chloride is 14.8 kJ/mol. Determine the lowest temperature reached during the original formulation.

Answer 1

Answer: Tfinal = 7.1°C

Explanation:

heat released or absorbed = mass × specific heat capacity × change in temperature

q = m × cg × ΔT  (eqn 1)

Note:  ΔT = (Tfinal - Tinitial)

(q = ? ΔHsoln=25.7kJ/mole = 25700J/mole; mass of solution, m = 100 + 35g = 135g; cg = specific heat capacity of water  = 4.18 J°C-1g-1; ΔT = ? masss of solute, NH4NO3 = 35g, molar mass of solute, NH4NO3 = 80g)

molar enthalpy of solution, ΔHsoln = heat absorbed or released ÷ moles of solute, n

ΔHsoln = q ÷ n

q = ΔHsoln × n

moles solute, n = mass solute (g) ÷ molar mass solute (g mol-1)

moles of solute, n = 35g/80g/mol = 0.4375 moles

q = 25700J/mol × 0.4375 mol = 11243.75J

From equation 1 above, ΔT = q / (m × cg) = 11243.75J / (135 × 4.18 J°C-1g-1) = 19.9°C

Since the reaction is endothermic, Tinitial > Tfinal, therefore, Tfinal = Tinitial - ΔT = 27 - 19.9 = 7.1°C