NH4NO3(s)+H2O(l)→NH4NO3(aq) ΔH = +25.7 kJ.
What is the final temperature in a squeezed cold pack that contains 49.0g of NH4NO3 dissolved in 125 mL of water? Assume a specific heat of 4.18J/(g⋅∘C) for the solution, an initial temperature of 28.5∘C, and no heat transfer between the cold pack and the environment.
Please help. I keep getting an increase in heat, which I know isn't right. But I can't seem to figure out where I went wrong. Please explain! Thanks!
Instead of me guessing what you did wrong why don't you post your work and let me find the error?
To find the final temperature of the solution, we can use the equation:
q = mcΔT
Where:
q = heat transferred (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)
First, let's calculate the mass of the solution. Since we have 49.0 g of NH4NO3 dissolved in 125 mL of water, we need to convert mL to grams. The density of water is approximately 1 g/mL, so:
Mass of water = Volume of water × Density of water
Mass of water = 125 mL × 1 g/mL
Mass of water = 125 g
Now, let's calculate the total mass of the solution:
Mass of solution = Mass of NH4NO3 + Mass of water
Mass of solution = 49.0 g + 125 g
Mass of solution = 174.0 g
Next, we calculate the heat transferred (q). We know that the reaction is endothermic, so the heat is gained by the solution. Therefore, q is positive and can be calculated as follows:
q = ΔH (enthalpy change) = +25.7 kJ = +25.7 × 1000 J (since 1 kJ = 1000 J)
Now, we can rearrange the equation and solve for ΔT:
ΔT = q / (m × c)
ΔT = (+25.7 × 1000 J) / (174.0 g × 4.18 J/(g⋅∘C))
Calculating ΔT:
ΔT = 147.94 ∘C
Finally, we can calculate the final temperature by adding the change in temperature (ΔT) to the initial temperature:
Final temperature = Initial temperature + ΔT
Final temperature = 28.5 ∘C + 147.94 ∘C
Final temperature ≈ 176.44 ∘C
Therefore, the final temperature in the squeezed cold pack would be approximately 176.44 ∘C.
To solve this problem, we need to use the equation:
q = mcΔT
where q is the heat absorbed or released by the solution, m is the mass of the solution, c is the specific heat capacity of the solution, and ΔT is the change in temperature.
First, let's calculate q for the dissolution of NH4NO3:
q = ΔH × n
where ΔH is the enthalpy change of the dissolution reaction, and n is the number of moles of NH4NO3 dissolved.
To find n, we need to calculate the number of moles of NH4NO3:
moles = mass / molar mass
The molar mass of NH4NO3 is:
M(N) = 14.01 g/mol
M(H) = 1.01 g/mol
M(O) = 16.00 g/mol
so,
M(NH4NO3) = (4 × M(H)) + M(N) + (3 × M(O))
= (4 × 1.01 g/mol) + 14.01 g/mol + (3 × 16.00 g/mol)
= 80.05 g/mol
moles = 49.0 g / 80.05 g/mol
Now, let's calculate q:
q = ΔH × n
= (+25.7 kJ) × (moles)
Since the density of the solution is given, we can convert the volume from mL to grams:
density = mass / volume
mass = density × volume
For water, the density is approximately 1.00 g/mL, so:
mass = (1.00 g/mL) × 125 mL
Next, let's calculate the total mass of the solution:
mass = mass of NH4NO3 + mass of water
Finally, let's calculate the change in temperature:
q = mcΔT
Solving for ΔT:
ΔT = q / (mc)
Plug in the values we have calculated and solve for ΔT.
ΔT will give us the change in temperature, which we add to the initial temperature to get the final temperature.
That's it! This step-by-step explanation should help you identify where you might have gone wrong in your calculation.