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 following endothermic reaction.

NH4NO3(s) + H2O(l) = NH4NO3(aq) ÄH = +25.7 kJ

What is the final temperature in a squeezed cold pack that contains 50.0 g of NH4NO3 dissolved in 115 mL of water? Assume a specific heat of 4.18 J/(g·°C) for the solution, an initial temperature of 25.0°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. Hint: The process takes place at constant pressure.

So i know at constant pressure i have to use

Cs * M of solution * deltaT = -(molesNH4NO3)*DeltaH of reaction

Can someone please set this problem up because I tried to set it up but had difficulties with some of the values.

Answer:

Cs * M of solution * deltaT = -(molesNH4NO3)*DeltaH of reaction

4.18 J/(g·°C) * (50.0 g + 115 mL * 1.0 g/mL) * deltaT = -(50.0 g/80.04 g/mol)*25.7 kJ

deltaT = -25.7 kJ / (4.18 J/(g·°C) * (50.0 g + 115 mL * 1.0 g/mL))

deltaT = -25.7 kJ / (4.18 J/(g·°C) * (165.0 g))

deltaT = -25.7 kJ / 690.7 J/°C

deltaT = -37.1°C

Final temperature = 25.0°C - 37.1°C = -12.1°C

Sure! Let's set up the problem step-by-step.

Step 1: Calculate the mass of water in the solution.
Given:
Density of water = 1.0 g/mL
Volume of water (V) = 115 mL

Mass of water (m) = Density of water × Volume of water
m = 1.0 g/mL × 115 mL
m = 115.0 g

Step 2: Calculate the moles of NH4NO3.
Given:
Mass of NH4NO3 (M) = 50.0 g
Molar mass of NH4NO3 = 80.04 g/mol

Moles of NH4NO3 = Mass of NH4NO3 / Molar mass of NH4NO3
molesNH4NO3 = 50.0 g / 80.04 g/mol
molesNH4NO3 ≈ 0.6247 mol (rounded to 4 decimal places)

Step 3: Calculate the heat absorbed by the solution.
Given:
Specific heat capacity of the solution (Cs) = 4.18 J/(g·°C)
Initial temperature (Ti) = 25.0°C
Final temperature (Tf) = ?

Heat absorbed = (molesNH4NO3 × ΔH of reaction) + (Cs × mass of solution × ΔT)

ΔH of reaction = +25.7 kJ = +25.7 × 10^3 J
mass of solution = mass of water + mass of NH4NO3

mass of solution = m + Mass of NH4NO3
mass of solution = 115.0 g + 50.0 g
mass of solution = 165.0 g

ΔT = Tf - Ti

Now let's plug in the values and solve for Tf:
(molesNH4NO3 × ΔH of reaction) + (Cs × mass of solution × ΔT) = 0

(0.6247 mol × 25.7 × 10^3 J/mol) + (4.18 J/(g·°C) × 165.0 g × (Tf - 25.0°C)) = 0

Solve the equation for Tf to find the final temperature.

To solve this problem, let's break it down step by step:

Step 1: Convert the given volume of water to mass.
Given:
- Density of water = 1.0 g/mL.
- Volume of water = 115 mL.

To find the mass of water, we can use the formula: mass = volume x density.
Mass of water = 115 mL x 1.0 g/mL = 115 g.

Step 2: Calculate the moles of NH4NO3.
Given:
- Mass of NH4NO3 = 50.0 g.
- Molar mass of NH4NO3 = 80.04 g/mol.

To find the moles of NH4NO3, we can use the formula: moles = mass / molar mass.
Moles of NH4NO3 = 50.0 g / 80.04 g/mol = 0.624 mol.

Step 3: Calculate the change in temperature.
Given:
- Initial temperature = 25.0 °C.

To find the change in temperature, we can subtract the initial temperature from the final temperature.
Change in temperature = Final temperature - Initial temperature.

We need to rearrange the formula Cs * M of solution * deltaT = -(molesNH4NO3) * DeltaH of reaction to solve for deltaT.

Rearranging: deltaT = -(molesNH4NO3 * DeltaH of reaction) / (Cs * M of solution).

Step 4: Calculate the specific heat (Cs) of the solution.
Given:
- Specific heat of the solution = 4.18 J/(g·°C).

Step 5: Calculate the final temperature.
Given:
- DeltaH of reaction = +25.7 kJ (to convert to J, multiply by 1000).

deltaT = -((0.624 mol) * (25.7 kJ * 1000 J/kJ)) / ((4.18 J/(g·°C)) * (115 g + 50.0 g))
deltaT = -16046.4 J / (4805.7 J/(g·°C))
deltaT = -3.34 °C

Final temperature = Initial temperature + deltaT
Final temperature = 25.0 °C - 3.34 °C = 21.7 °C

Therefore, the final temperature in a squeezed cold pack that contains 50.0 g of NH4NO3 dissolved in 115 mL of water is approximately 21.7 °C.