A) if you added 8g of ammonium nitrate to 25mL of room temperature water, what would be the temperature in °C?
You need to look up the delta H solution of NH4NO3. It probably will be listedin kJ/mol. Convert 8g to mols and multply by delta H to find q for this problem.
Then -q = (mass H2O X specific heat H2O x (Tfinal-Tinitial)
Solve for Tf
To determine the resulting temperature after adding ammonium nitrate to water, we need to consider the heat transfer that occurs during the dissolution process. This can be calculated using the equation:
q = m × c × ΔT
Where:
q is the amount of 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), and
ΔT is the change in temperature (in °C).
To solve this problem, we need to know the specific heat capacity of the solution and the enthalpy change (ΔH) associated with dissolving ammonium nitrate in water. The specific heat capacity of water is approximately 4.18 J/g°C.
However, since the question only provides the mass of ammonium nitrate (8g) and the volume of water (25mL), we need to convert the volume to mass and combine it with the mass of ammonium nitrate to calculate the resulting temperature.
To convert the volume of water to mass, we can use the density of water. The density of water at room temperature is approximately 1g/mL. Therefore, the mass of 25mL of water would be:
Mass of water = Volume × Density = 25mL × 1g/mL = 25g
Now we have the total mass of the solution, which is the sum of the mass of ammonium nitrate and water:
Total mass = Mass of ammonium nitrate + Mass of water = 8g + 25g = 33g
We can plug these values into the heat transfer equation to solve for the change in temperature:
q = m × c × ΔT
ΔT = q / (m × c)
Now, to find the heat transferred (q), we can use the enthalpy change of dissolving ammonium nitrate in water. The enthalpy change for the dissolution of ammonium nitrate is exothermic and releases 25.7 kJ/mol of heat.
To calculate the heat transferred (q), we need to convert the mass of ammonium nitrate to moles and then use the enthalpy change:
Molar mass of ammonium nitrate = 80 g/mol
Moles of ammonium nitrate = Mass of ammonium nitrate / Molar mass of ammonium nitrate = 8g / 80 g/mol = 0.1 mol
Now we can calculate the heat transferred:
q = ΔH × moles = 25.7 kJ/mol × 0.1 mol = 2.57 kJ
Since 1 kJ (kilojoule) is equal to 1000 J, we need to convert our value to Joules:
q = 2.57 kJ × 1000 J/kJ = 2570 J
Finally, we can substitute the values into the equation for ΔT:
ΔT = q / (m × c) = 2570 J / (33g × 4.18 J/g°C) ≈ 19.6°C
Therefore, adding 8g of ammonium nitrate to 25mL of room temperature water would result in a temperature increase of approximately 19.6°C.