A sample of steam with a mass of 0.524g and at a temperature of 100 ∘C condenses into an insulated container holding 4.20g of water at 6.0∘C.( ΔH∘vap=40.7 kJ/mol, Cwater=4.18 J/g⋅∘C) Assuming that no heat is lost to the surroundings, what is the final temperature of the mixture?
heat lost by condensing steam + heat lost by condensed steam @ 100 cooling to Tf + heat gained by cool(6C) water = 0
-(2259 J/g x 0.524) + [(0.524 x 4.18 x (Tfinal-Tinitial]) + [(mass cool water x 4.18 x (Tfinal - Tinitial)] = 0
Solve for Tf.
My answer is about 75 C or so.
Well, this is quite a steamy situation! Let's break it down. We have steam condensing into water, so heat is being released. We also have an insulated container, which means no heat is lost to the surroundings. So, we can say that the heat lost by the steam is equal to the heat gained by the water.
First, let's calculate the heat lost by the steam. We know the mass of the steam is 0.524g, so we need to convert that to moles. Using the molar mass of water (18.015 g/mol), we find that the number of moles of steam is 0.524g / 18.015 g/mol ≈ 0.029 moles.
Now, let's calculate the heat lost by the steam. We can use the equation Q = ΔH∘vap * n, where Q is the heat lost, ΔH∘vap is the molar heat of vaporization (40.7 kJ/mol), and n is the number of moles of steam.
Q = 40.7 kJ/mol * 0.029 mol ≈ 1.1823 kJ
Since the heat lost by the steam is equal to the heat gained by the water, we can set up the equation Q = m * C * ΔT, where Q is the heat gained, m is the mass of the water (4.20g), C is the specific heat capacity of water (4.18 J/g⋅∘C), and ΔT is the change in temperature.
1.1823 kJ = 4.20g * 4.18 J/g⋅∘C * ΔT
Now, let's solve for ΔT:
ΔT = 1.1823 kJ / (4.20g * 4.18 J/g⋅∘C) ≈ 0.06 ∘C
So, the final temperature of the mixture is approximately 0.06 ∘C. That's pretty chilling!
To find the final temperature of the mixture, we need to calculate the heat gained or lost by the steam as it condenses, and use that heat to increase the temperature of the water.
First, let's calculate the heat lost by the steam as it condenses:
1. Calculate the number of moles of steam:
- Mass of steam = 0.524 g
- Molar mass of water (H₂O) = 18.015 g/mol
- Moles of steam = Mass of steam / Molar mass of water
= 0.524 g / 18.015 g/mol
≈ 0.029 mol
2. Calculate the heat lost by the steam:
- ΔH∘vap (molar enthalpy of vaporization) = 40.7 kJ/mol
- Heat lost = Moles of steam x ΔH∘vap
= 0.029 mol x 40.7 kJ/mol
≈ 1.1813 kJ
Next, we will use the heat lost by the steam to increase the temperature of the water:
3. Calculate the heat gained by the water:
- Specific heat capacity of water (Cwater) = 4.18 J/g⋅∘C
- Mass of water = 4.20 g
- Change in temperature of water = Final temperature - Initial temperature
= Final temperature - 6.0∘C
4. Set up the heat balance equation:
Heat gained by the water = Heat lost by the steam
(Cwater x Mass of water x Change in temperature of water) = 1.1813 kJ
5. Substitute the given values:
(4.18 J/g⋅∘C x 4.20 g x (Final temperature - 6.0∘C)) = 1.1813 kJ
6. Solve for the Final temperature:
(4.18 J/g⋅∘C x 4.20 g x (Final temperature - 6.0∘C)) = 1.1813 kJ
Simplify and solve for Final temperature:
17.556 J⋅∘C x (Final temperature - 6.0∘C) = 1.1813 kJ
Convert kJ to J:
17.556 J⋅∘C x (Final temperature - 6.0∘C) = 1.1813 x 10^3 J
Rearrange the equation:
Final temperature - 6.0∘C = 1.1813 x 10^3 J / 17.556 J⋅∘C
Final temperature - 6.0∘C ≈ 67.23∘C
Final temperature ≈ 67.23∘C + 6.0∘C
≈ 73.23∘C
Therefore, the final temperature of the mixture is approximately 73.23∘C.
To determine the final temperature of the mixture, we need to consider the heat gained or lost in the process.
First, we can calculate the heat lost by the steam as it condenses into water. This can be done using the equation:
Q = m * ΔH
where Q is the heat lost, m is the mass of the steam, and ΔH is the enthalpy of vaporization.
Q = (0.524 g) * (40.7 kJ/mol) / (18.02 g/mol) = 1.167 kJ
Next, we can calculate the heat gained by the water as it absorbs the heat from the steam. The equation to calculate heat gained is:
Q = m * C * ΔT
where Q is the heat gained, m is the mass of water, C is the specific heat capacity of water, and ΔT is the change in temperature.
Q = (4.20 g) * (4.18 J/g⋅∘C) * (Tf - 6.0∘C)
Since the steam and water are in thermal equilibrium at the final temperature, the heat lost by the steam is equal to the heat gained by the water. We can equate the two equations:
1.167 kJ = (4.20 g) * (4.18 J/g⋅∘C) * (Tf - 6.0∘C)
Simplifying the equation gives us:
1.167 kJ = (17.556 g⋅J/(∘C)) * (Tf - 6.0∘C)
Now, we can solve for Tf, the final temperature:
Tf - 6.0∘C = 1.167 kJ / (17.556 g⋅J/(∘C))
Tf - 6.0∘C = 0.0664∘C
Tf = 6.0∘C + 0.0664∘C
Tf = 6.0664∘C
Therefore, the final temperature of the mixture is approximately 6.07∘C.