What is the final temperature in a squeezed cold pack that contains 45.0 g of NH4NO3 dissolved in 125 mL of water?

[匿名]

2018-01-17 10:33 am

Assume a specific heat of 4.18J/(g⋅∘C) for the solution, an initial temperature of 28.0 ∘C, and no heat transfer between the cold pack and the environment.

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) ΔH = +25.7 kJ.

回答 (1)

wanszeto

2018-01-17 10:52 am

✔ 最佳答案

Let T°C be the final temperature.

Molar mass of NH₄NO₃ = (14.0×2 + 1.0×4 + 16.0×3) g/mol = 80.0 g/mol
No. of moles of NH₄NO₃ used = (45.0 g) / (80.0 g/mol) = 45.0/80.0 mol

NH₄NO₃(s) + aq → NH₄NO₃(aq) …… ΔH = +25.7 kJ
Dissolving of 1 mole of NH₄NO₃ absorbs 25.7 kJ of heat.
Heat absorbed in reaction = (25.7 kJ/mol) × (45.0/80.0 mol) × (1000 J/kJ)

Density of water = (1.00 g/mL)
Mass of water in the solution = (125 mL) × (1.00 g/mL) = 125 g
Mass of the solution, m = (125 + 45.0) g = 170 g
Heat released by the solution = m c ΔT = (170 g) × (4.18 J/g°) × [(28.0 - T)°C]

Heat released by the solution = Heat absorbed in reaction
170 × 4.18 × (28.0 - T) = 25.7 × (45.0/80.0) × 1000
28.0 - T = [25.7 × (45.0/80.0) × 1000 / (170 × 4.18)]
28.0 - T = 20.3
T = 7.7

Final temperature = 7.7°C

👍 3 👎 0