Instant cold packs used to treat athletic injuries contain solid NH_4NO_3 and a
ID: 477021 • Letter: I
Question
Instant cold packs used to treat athletic injuries contain solid NH_4NO_3 and a pouch of water. When the pack is squeezed, the pouch breaks and the solid dissolves lowering the temperature because of the following endothermic reaction NH_4NO_3(s) + H_20(l) rightarrow NH_4NO_3(aq) delta H = +25.7 kJ What is the final temperature in a squeezed cold pack that contains 40.0 g of NH_4NO_3 dissolved in 125 mL of water? Assume a specific heat of 4.18 J/(g. degree C) for the solution, an initial temperature of 25.0 degree C, and no heat transfer between the cold pack and the environment. To find the mass of water use the density of water = 1.0 g/mL. When a 4.25-g sample of solid ammonium nitrate dissolves n 60.0 g of water in a coffee-cup calorimeter (see figure below), the temperature drops from 23.0 degree C to 16.1 degree C. Calculate delta H (in kJ/mol NH_4NO_3) for the solution process shown below. Assume that the specific heat of the solution is the same as that of pure water. NH_4NO_3(s) rightarrow NH_4^+ (aq) + NO_3^-(aq) Is this process endothermic or exothermic? Consider a glass of 207 mL of water at 27 degree C. Calculate the mass of ice at 15 degree C that must be added to cool the water to 10 degree C after thermal equilibrium is achieved. To find the mass of water use the density of water 1.0 g/mL.Explanation / Answer
When a 4.25-g sample of solid ammonium nitrate dissolves in 60.0 g of water in a coffee-cup calorimeter,?
the temperature drops from 23.0 C to 16.1 C.
a) Calculate H (in kJ/mol NH4NO3) for the solution process
NH4NO3(s)NH+4(aq)+NO3(aq)
Assume that the specific heat of the solution is the same as that of pure water.
Express your answer to two significant figures and include the appropriate units.
b) Is this process endothermic or exothermic?
= (4.25 g NH4NO3) / (80.04352 g NH4NO3/mol) = 0.053096 mol NH4NO3
(4.18 J/g·°C) x (60.0 g) x (23.0 - 16.1)°C = 1730.52 J
(1730.52 J) / (0.053096 mol NH4NO3) = 32592.285 J/mol = 32.5 kJ/mol NH4NO3
b) endothermic
And so H is positive.
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What is the final temperature in a squeezed cold pack that contains 40 g of NH4NO3 dissolved in 125 mL of water? Assume a specific heat of 4.18J/g C for the solution, an initial temperature of 27.5 C, and no heat transfer between the cold pack and the environment.
ALSO:
NH4NO3 (s) + H2O(l) -->NH4NO3 (aq) Delta H = +25.7 kJ.
find moles, using molar mass:
40 g of NH4NO3 / 80.043 g/mol = 0.499 moles NH4NO3
find dH by dissolving 0.499 moles NH4NO3:
0.581 moles NH4NO3 @ +25.7 kJ / mol = + 12.824kJ absorbed by the reaction
+ 12.824kJ absorbed by the reaction
- 12.824kJ lost by the water
-12.824Joules by the water
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using the density of water @ 1 gram/ml, the 125 ml H2O gass a mass of 125 grams...
that 125 grams , along with the 40 g NH4NO3 together weighs 165 grams
find the dTemp for 165 grams of solution,
assuming that we can use water's specific heat:of 4.184,
dH = m C dT
- 12824 J = 165 g (4.184 J/g.C) dT
dT = - 18.57 C
What is the final temperature
27.5 - 20.8 = 6.7 Canswer
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