The heat of solution of ammonium chloride is 15.2 kJ/mol. If a 6.187 g sample of NH4Cl is added to 65.0 mL of water in a calorimeter at 26.5°C, what is the minimum temperature reached by the solution? [specific heat of water = 4.18 J/g°C; heat capacity of the calorimeter = 365. J/°C]
The sum of heats gained is zero.
heatgainedwater+heataddedtoNH3Cl=0
65g*cwater*(Tf-26.5)+6.187/molmass*15.2kj/mol=0
now, change c water to Kj/g to make units work out (kJ)
cwater=.004kJ/gC
and that ought to let you work out what is tf.
To find the minimum temperature reached by the solution, we can use the principle of heat transfer.
The equation for heat transfer is:
Q = mcΔT
Where:
Q = heat transferred (in Joules)
m = mass of the substance (in grams)
c = specific heat capacity of the substance (in J/g°C)
ΔT = change in temperature (in °C)
First, we need to calculate the amount of heat transferred when the ammonium chloride dissolves in water:
Q1 = msΔT
Where:
Q1 = heat transferred when NH4Cl dissolves (in Joules)
m = mass of NH4Cl (in grams)
s = heat of solution of NH4Cl (in J/g)
ΔT = change in temperature (in °C)
Given:
m = 6.187 g
s = 15.2 kJ/mol
To convert kJ/mol to J/g, divide by the molar mass of NH4Cl:
Molar mass of NH4Cl = 14.01 g/mol (N) + 4(1.01 g/mol) (H) + 35.45 g/mol (Cl)
Molar mass of NH4Cl = 53.49 g/mol
s = (15.2 kJ/mol) / (53.49 g/mol) = 0.284 J/g
Next, we can calculate Q1:
Q1 = (6.187 g) * (0.284 J/g) * ΔT
Now let's calculate the second term of the equation, which is the heat transferred from the water and the calorimeter:
Q2 = mcΔT
Where:
Q2 = heat transferred from water and calorimeter (in Joules)
m = mass of water (in grams)
c = specific heat capacity of water (in J/g°C)
ΔT = change in temperature (in °C)
Given:
m = 65.0 mL = 65.0 g
c = 4.18 J/g°C
ΔT = final temperature - initial temperature = Tf - Ti
We are given the initial temperature (Ti) as 26.5°C. To solve for the final temperature (Tf), we need to rearrange the equation:
Tf = ΔT + Ti
Now we can calculate Q2:
Q2 = (65.0 g) * (4.18 J/g°C) * ΔT
Finally, since energy is conserved, the total heat transferred Q1 + Q2 = 0:
Q1 + Q2 = 0
Substituting the values, we have:
(6.187 g) * (0.284 J/g) * ΔT + (65.0 g) * (4.18 J/g°C) * ΔT = 0
Now we can solve for ΔT:
(6.187 g * 0.284 J/g + 65.0 g * 4.18 J/g°C) * ΔT = 0
(1.757108 g + 271.7 g°C) * ΔT = 0
273.457108 g°C * ΔT = 0
ΔT = 0 / 273.457108 g°C
ΔT = 0
Since ΔT = 0, the minimum temperature reached by the solution is 26.5°C, which is the initial temperature of the water in the calorimeter.
To find the minimum temperature reached by the solution, we first need to calculate the heat involved in the process. The heat involved in the process can be calculated using the equation:
q = m × C × ΔT
where:
q = heat involved (in J)
m = mass of the substance (in g)
C = specific heat of the substance (in J/g°C)
ΔT = change in temperature (in °C)
Step 1: Calculate the heat involved in dissolving the NH4Cl.
The heat involved in dissolving the NH4Cl can be calculated using the following equation:
q1 = m1 × ΔH1
where:
q1 = heat involved in dissolving NH4Cl (in J)
m1 = moles of NH4Cl
ΔH1 = heat of solution of NH4Cl (in J/mol)
First, we need to calculate the moles of NH4Cl using the molar mass of NH4Cl.
Molar mass of NH4Cl = 14.01 g/mol (N) + 4 × 1.01 g/mol (H) + 1 × 35.45 g/mol (Cl)
= 53.49 g/mol
moles of NH4Cl = mass of NH4Cl / molar mass of NH4Cl
= 6.187 g / 53.49 g/mol
= 0.1157 mol
Now, we can calculate the heat involved in dissolving NH4Cl:
q1 = m1 × ΔH1
= 0.1157 mol × 15.2 kJ/mol × 1000 J/kJ
= 1758 J (or 1.758 kJ)
Step 2: Calculate the heat involved in warming the water and the calorimeter.
The heat involved in warming the water and the calorimeter can be calculated using the equation:
q2 = (m2 + m3) × C × ΔT
where:
q2 = heat involved in warming the water and the calorimeter (in J)
m2 = mass of water (in g)
m3 = mass of calorimeter (in g)
C = specific heat of water (in J/g°C)
ΔT = change in temperature (in °C)
Given that the mass of water is 65.0 mL and the density of water is approximately 1 g/mL, we can calculate the mass of water as follows:
m2 = volume of water × density of water
= 65.0 mL × 1 g/mL
= 65.0 g
Now, we can calculate the heat involved in warming the water and the calorimeter using the given values:
q2 = (m2 + m3) × C × ΔT
= (65.0 g + m3) × 4.18 J/g°C × (final temperature - 26.5°C)
Step 3: Equation of heat conservation.
According to the principle of heat conservation, the heat lost by the NH4Cl solution (q1) is equal to the heat gained by the water and the calorimeter (q2).
q1 = q2
Substituting the calculated values:
1.758 kJ = (65.0 g + m3) × 4.18 J/g°C × (final temperature - 26.5°C)
Simplifying the equation:
1758 J = (65.0 g + m3) × 4.18 J/g°C × (final temperature - 26.5°C)
423.5 = (65.0 g + m3) × (final temperature - 26.5°C)
Step 4: Solve for the minimum temperature.
To find the minimum temperature reached by the solution, we need to find the value of (final temperature - 26.5°C) when q1 = q2.
423.5 = (65.0 g + m3) × (final temperature - 26.5°C)
Since the calorimeter's heat capacity is given as 365 J/°C, we can substitute m3 with 365 g:
423.5 = (65.0 g + 365 g) × (final temperature - 26.5°C)
423.5 = 430 g × (final temperature - 26.5°C)
Dividing both sides of the equation by 430 g:
423.5 / 430 = final temperature - 26.5°C
Final temperature = (423.5 / 430) + 26.5°C
By calculating this expression, you will find the minimum temperature reached by the solution.