The heat of solution of ammonium nitrate is 26.2 kJ/mol, and a 7.315 g sample of NH4NO3 is added to 40.0 mL of water in a constant pressure calorimeter initially at 25.2°C, what is the final temperature reached by the solution? (The specific heat of water = 4.184 J/g•°C)
Heat lost by KNO3 + heat gained by water = 0
q lost by NH4NO3 is
26.3 kJ x (7.315/80) = approx 2.5 kJ but you need a better answer.
Since dH is + you know the reaction is endothermic and the solution will get colder.
2.5 = mass H2O x specific heat H2O x delta T.
Solve for delta T and subtract from 25.2. Remember that 2.5 is approximate.
To determine the final temperature reached by the solution, we can use the principle of energy conservation. The heat gained by the water is equal to the heat lost by the ammonium nitrate.
First, let's determine the amount of heat gained by water using the formula:
q_water = m_water * c_water * ΔT
Where:
q_water is the heat gained by water
m_water is the mass of water
c_water is the specific heat of water
ΔT is the change in temperature
We know that the mass of water is 40.0 mL, which is equal to 40.0 g since the density of water is 1 g/mL. The specific heat of water is given as 4.184 J/g•°C. We need to calculate the change in temperature.
Next, let's find the amount of heat lost by the ammonium nitrate using the formula:
q_ammonium_nitrate = m_ammonium_nitrate * ΔH
Where:
q_ammonium_nitrate is the heat lost by the ammonium nitrate
m_ammonium_nitrate is the mass of ammonium nitrate
ΔH is the heat of solution of ammonium nitrate
We know that the mass of ammonium nitrate is 7.315 g, and the heat of solution of ammonium nitrate is 26.2 kJ/mol. To convert kJ to J, we multiply by 1000. We can then calculate the amount of heat lost by the ammonium nitrate.
Since energy is conserved, the heat gained by water should be equal to the heat lost by ammonium nitrate. Therefore, we can set the two equations equal to each other:
m_water * c_water * ΔT = m_ammonium_nitrate * ΔH
Now, we can rearrange the equation to solve for ΔT, the change in temperature, which is the same as the final temperature reached by the solution.
ΔT = (m_ammonium_nitrate * ΔH) / (m_water * c_water)
Plugging in the values will give us the final temperature reached by the solution.
To solve this question, we can use the equation:
q = m * c * ΔT
Where:
q is the heat absorbed or released by the system
m is the mass of the substance
c is the specific heat of the substance
ΔT is the change in temperature
First, let's calculate the heat absorbed or released by the system:
q = heat of solution = -26.2 kJ/mol
Let's convert the mass of NH4NO3 to moles:
molar mass of NH4NO3 = 80.043 g/mol
moles of NH4NO3 = 7.315 g / 80.043 g/mol = 0.0914 mol
Now, let's calculate the heat absorbed or released by the reaction:
q = -26.2 kJ/mol * 0.0914 mol = -2.39 kJ
Next, let's calculate the heat absorbed or released by the water:
q = m * c * ΔT
m = 40.0 mL = 40.0 g (since the density of water is approximately 1 g/mL)
c = 4.184 J/g•°C
ΔT = final temperature - initial temperature = ΔT
Now, let's rearrange the equation to solve for ΔT:
ΔT = q / (m * c) = -2.39 kJ / (40.0 g * 4.184 J/g•°C)
ΔT ≈ -14.3 °C (rounded to one decimal place)
Finally, let's calculate the final temperature:
final temperature = initial temperature + ΔT = 25.2°C + (-14.3°C)
final temperature ≈ 10.9 °C (rounded to one decimal place)
Therefore, the final temperature reached by the solution is approximately 10.9 °C.