NH4NO3(s)+H2O(l)→NH4NO3(aq) ΔH = +25.7 kJ.

What is the final temperature in a squeezed cold pack that contains 45.5g of NH4NO3 dissolved in 125 mL of water? Assume a specific heat of 4.18J/(g⋅∘C) for the solution, an initial temperature of 29.0∘C, and no heat transfer between the cold pack and the environment

To determine the final temperature of the solution, we need to calculate the heat absorbed by the system.

First, let's convert the mass of NH4NO3 from grams to moles:
molar mass of NH4NO3 = (14.01 g/mol of N) + (4 * 1.01 g/mol of H) + (3 * 16.00 g/mol of O) = 80.04 g/mol
moles of NH4NO3 = mass / molar mass = 45.5 g / 80.04 g/mol ≈ 0.5684 mol

Next, let's calculate the heat absorbed using the equation:
q = m * c * ΔT
where:
q = heat absorbed (in J)
m = mass of the solution (in g)
c = specific heat capacity of the solution (in J/g⋅°C)
ΔT = change in temperature (final temperature - initial temperature) (in °C)

mass of the solution = mass of NH4NO3 + mass of water
mass of water = volume of water * density of water
density of water = 1 g/mL

mass of water = 125 mL * 1 g/mL = 125 g
mass of the solution = 45.5 g + 125 g = 170.5 g

ΔT = final temperature - initial temperature

Now, let's rearrange the equation to solve for the final temperature:
ΔT = q / (m * c)
final temperature = ΔT + initial temperature

Using the given values:
m = 170.5 g
c = 4.18 J/(g⋅°C)
initial temperature = 29.0°C

Now, let's calculate q:
q = ΔH = +25.7 kJ = +25.7 * 10^3 J

ΔT = q / (m * c)
ΔT = (25.7 * 10^3 J) / (170.5 g * 4.18 J/(g⋅°C))
ΔT ≈ 37.24 °C

Finally, substitute the values into the final temperature equation:
final temperature = ΔT + initial temperature
final temperature = 37.24 °C + 29.0 °C
final temperature ≈ 66.24 °C

Therefore, the final temperature in the squeezed cold pack is approximately 66.24°C.

To find the final temperature in the cold pack, we need to use the equation for heat transfer:

q = m * c * ΔT

where:
q is the heat transferred (in joules),
m is the mass of the solution (in grams),
c is the specific heat capacity of the solution (in J/(g⋅∘C)),
ΔT is the change in temperature (final temperature - initial temperature).

In this case, the heat transferred (q) is equal to the enthalpy change (ΔH) of the dissolution process of NH4NO3.

First, let's convert the mass of NH4NO3 from grams to moles using its molar mass. The molar mass of NH4NO3 is 80.04 g/mol:

moles of NH4NO3 = mass of NH4NO3 (g) / molar mass of NH4NO3 (g/mol)

moles of NH4NO3 = 45.5 g / 80.04 g/mol

Next, we need to calculate the heat transferred (q) using the enthalpy change (ΔH) and the number of moles of NH4NO3:

q = ΔH * moles of NH4NO3

Since the enthalpy change (ΔH) is given as +25.7 kJ, we need to convert it to joules for consistency:

ΔH = +25.7 kJ * 1000 J/1 kJ

Now we can calculate q:

q = (25.7 kJ * 1000 J/1 kJ) * (moles of NH4NO3)

Next, we need to calculate the final volume of the solution. The volume of the solution is given as 125 mL, which we can convert to liters:

volume of solution = 125 mL / 1000 mL/L

Now, we need to calculate the specific heat capacity of the solution. The specific heat capacity (c) is given as 4.18 J/(g⋅∘C).

Next, we rearrange the heat transfer equation to solve for the change in temperature:

ΔT = q / (m * c)

Finally, we can substitute the known values into the equation:

ΔT = (q) / (mass of solution) * (specific heat capacity)

Now that we have the change in temperature, we can calculate the final temperature by adding the change in temperature to the initial temperature:

Final temperature = Initial temperature + ΔT

By following these steps and plugging in the given values, you should be able to find the final temperature in the cold pack.

First, determine q for the reaction. That is

25700 J x (45.5/molar mass NH4NO3) = approx 15,000 J.
Then
q = [mass H2O x specific heat H2O x (Tfinal-Tinitial)
q from above
mass H2O = 125g
specific heat H2O given
Solve for Tf
Ti is given