NH4NO3(s) + H2O(l) NH4NO3(aq) ΔH = +25.7 kJ
What is the final temperature in a squeezed cold pack that contains 40.0 g of NH4NO3 dissolved in 125 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.
For this question im suppose to use the q=mc delta T ?
That 25.7 kJ usually is quoted as 235.7 kJ/mol which means 25.7 kJ/80g.
Therefore, heat absorbed by the NH4NO3 will be 25,700 J x (40/80) = about 13,000 J
Then -13000J = 125 x 4.184 x (Tf-Ti)
Tf = unknown
Ti = 25C
Tf is something like 0.5C but that's an estimate.
Yes, you are correct. In order to find the final temperature using the equation q = mcΔT, where q is the heat transferred, m is the mass of the solution, c is the specific heat capacity of the solution, and ΔT is the change in temperature, you need to plug in the appropriate values into the equation. Here's how you can calculate the final temperature:
1. Calculate the heat transferred using the equation q = mcΔT:
q = (mass of solution) x (specific heat capacity) x (ΔT)
Given:
Mass of solution = 40.0 g
Specific heat capacity of the solution = 4.18 J/(g·°C)
Initial temperature = 25.0°C
ΔT = final temperature - initial temperature
2. Rearrange the equation to solve for the final temperature (Tf):
Tf = (q / (m x c)) + Ti
Substitute the given values:
Tf = (q / (40.0 g x 4.18 J/(g·°C))) + 25.0°C
3. Calculate the heat transferred (q) using the enthalpy change (∆H) and the amount of substance (moles) present:
q = ∆H x (moles of NH4NO3)
Given:
Mass of NH4NO3 = 40.0 g
Molar mass of NH4NO3 = (14.01 g/mol + 1.01 g/mol + 14.01 g/mol + 3(16.00 g/mol)) = 80.04 g/mol
Calculate the moles of NH4NO3:
Moles of NH4NO3 = (40.0 g / 80.04 g/mol)
Substitute the value of moles in the equation:
q = 25.7 kJ x (40.0 g / 80.04 g/mol)
Convert kJ to J (1 kJ = 1000 J):
q = 25.7 kJ x 1000 J/1 kJ
4. Substitute the calculated value of q into the equation for Tf and solve for the final temperature:
Tf = ((25.7 kJ x 1000 J/1 kJ) / (40.0 g x 4.18 J/(g·°C))) + 25.0°C
Calculate Tf to find the final temperature.
Note: Remember to use consistent units throughout the calculations.
Yes, you are correct. To solve this question, you can use the equation q = mcΔT, where q is the heat energy transferred, 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 find the heat energy transferred in the dissolution of NH4NO3.
Given:
- ΔH = +25.7 kJ (the heat of solution)
- Mass of NH4NO3 = 40.0 g
- Specific heat capacity of the solution (c) = 4.18 J/(g·°C)
- Initial temperature (Ti) = 25.0°C
To find the heat energy (q) transferred, we can use the equation:
q = ΔH * m
Converting ΔH to joules:
ΔH = +25.7 kJ * 1000 J/1 kJ = +25,700 J
Plugging in the known values:
q = +25,700 J * 40.0 g = +1,028,000 J
Now that we have the amount of heat energy transferred, we can calculate the change in temperature (ΔT) using the equation:
q = mcΔT
Plugging in the known values:
+1,028,000 J = (40.0 g + 125 mL) * 4.18 J/(g·°C) * ΔT
Since the volume is given in milliliters (mL), we need to convert it to grams (g) before proceeding with the calculations.
The density of water is approximately 1 g/mL, so the mass of water (m) is:
m = 125 mL * 1 g/mL = 125 g
Now we can proceed with the calculations:
+1,028,000 J = (40.0 g + 125 g) * 4.18 J/(g·°C) * ΔT
Simplifying:
+1,028,000 J = 165 g * 4.18 J/(g·°C) * ΔT
Dividing both sides of the equation by (165 g * 4.18 J/(g·°C)):
+1,028,000 J / (165 g * 4.18 J/(g·°C)) = ΔT
Calculating ΔT:
ΔT ≈ 3.28 °C
Finally, to find the final temperature (Tf), we can add the change in temperature (ΔT) to the initial temperature (Ti):
Tf = Ti + ΔT = 25.0 °C + 3.28 °C
Tf ≈ 28.28 °C
Therefore, the final temperature in the squeezed cold pack is approximately 28.28 °C.