"Calculate the enthalpy of vaporization of water given that, when 5.00 grams of steam (gaseous H2O) at 100°C was introduced into a beaker containing 500 grams of water at 20.0°C, the temperature rose to 26.2°C."
q_H2O(l) + q_H2O(g)= E
nΔH_vap + m_gasCΔT + m_H2O(l)CΔT = E
(500g * 4.18J/g°C * (26.2°C-20°C)) + [(500g/18g/mol) * (40.7kJ/mol * 1000J)] + (5g * 2.0J/g°C * (26.2°C-100°C)) = E ???
Not sure if this is the right way, but help is much appreciated.
Well, I have to say, enthalpy calculations can sometimes make me steam with confusion! But don't worry, I'm here to help you out.
To calculate the enthalpy of vaporization (ΔH_vap) of water, we need to use the equation:
q_H2O(l) + q_H2O(g) = E
Where q_H2O(l) is the heat gained by the liquid water, q_H2O(g) is the heat gained by the gaseous water (steam), and E is the total heat gained.
Using the given information, we can break down the equation further:
m_H2O(l)CΔT + m_H2O(g)CΔT + nΔH_vap = E
Here, m_H2O(l) is the mass of liquid water, C is the specific heat capacity of water, ΔT is the change in temperature, m_H2O(g) is the mass of gaseous water (steam), n is the number of moles of water molecules (steam), and ΔH_vap is the enthalpy of vaporization of water.
Plugging in the values:
(500g * 4.18J/g°C * (26.2°C-20°C)) + (5g * 2.0J/g°C * (26.2°C-100°C)) + (n * ΔH_vap) = E
However, it seems you're using the molar enthalpy of vaporization (40.7 kJ/mol) instead of the specific enthalpy of vaporization. We need to convert it to J/g:
40.7 kJ/mol * (1 mol/18g) = 2261.1 J/g
So the corrected equation becomes:
(500g * 4.18J/g°C * (26.2°C-20°C)) + (5g * 2.0J/g°C * (26.2°C-100°C)) + (n * 2261.1 J/g) = E
Now, all that's left is to solve for ΔH_vap. I hope this clarifies things for you!
The equation you provided is almost correct, but there are a few modifications needed. Here is the corrected equation:
q_H2O(l) + q_H2O(g) = E
m_H2O(l) * C_H2O(l) * ΔT + m_H2O(g) * C_H2O(g) * ΔT + n_H2O(g) * ΔH_vap = E
Now let's plug in the given values:
m_H2O(l) = 500 g
C_H2O(l) = 4.18 J/g°C
ΔT = (26.2 - 20.0) °C = 6.2 °C
m_H2O(g) is the mass of gaseous water (steam) introduced, which is 5.00 g.
C_H2O(g) is the specific heat capacity of gaseous water, which is usually taken as 2.0 J/g°C.
n_H2O(g) is the number of moles of gaseous water, and it can be calculated using the molar mass of water (18 g/mol) and the mass of steam introduced (5.00 g).
n_H2O(g) = 5.00 g / 18 g/mol ≈ 0.2778 mol
Now you can calculate the enthalpy of vaporization ΔH_vap.
Using the equation:
(500 g * 4.18 J/g°C * (26.2°C-20.0°C)) + [(5.00 g * 2.0 J/g°C * (26.2°C-100°C))] + [(0.2778 mol * ΔH_vap)]
Remember that the units need to be consistent, so if you use kJ/mol for ΔH_vap, you should convert the other terms to kJ as well.
I hope this helps! Let me know if you have any further questions.
To calculate the enthalpy of vaporization of water, you will need to use the equation:
q_H2O(l) + q_H2O(g) = E
where q_H2O(l) represents the heat absorbed by the water at the lower temperature, q_H2O(g) represents the heat absorbed by the steam, and E represents the total heat absorbed by the system.
Now let's break down the equation step by step.
1. Calculate the heat absorbed by the water at the lower temperature (q_H2O(l)):
q_H2O(l) = m_H2O(l) * C * ΔT
where m_H2O(l) is the mass of water, C is the specific heat capacity of water, and ΔT is the temperature change.
Given:
m_H2O(l) = 500 g
C = 4.18 J/g°C
ΔT = 26.2°C - 20.0°C
q_H2O(l) = 500 g * 4.18 J/g°C * (26.2°C - 20.0°C)
2. Calculate the heat absorbed by the steam (q_H2O(g)):
q_H2O(g) = n * ΔH_vap
where n is the number of moles of water vaporized and ΔH_vap is the enthalpy of vaporization of water.
To find n, we can use the ideal gas law:
PV = nRT
Given:
P = pressure (not provided)
V = volume (not provided)
R = gas constant (R = 0.0821 atm∙L/mol∙K)
T = temperature (T = 373.2 K / 100°C)
You would need to know the pressure and volume to use the ideal gas law and find n. Unfortunately, the information given in the question is insufficient to calculate q_H2O(g).
3. Finally, calculate the total heat absorbed by the system (E):
E = q_H2O(l) + q_H2O(g)
At this point, you can substitute the values you have calculated for q_H2O(l) and q_H2O(g) into the equation to find E. However, since you do not have the value for q_H2O(g), you cannot determine the enthalpy of vaporization of water using the given information alone.
To accurately calculate the enthalpy of vaporization, you would need additional information such as pressure or volume.
The first term you have written is correct; the second and third are not.
The first term correctly identifies the energy needed to raise T of 500 g H2O from 20 to 26.2. Your third term should be the steam at 100 C condenses to liquid water at 100 C and helps heat the water to 26.2. That is
[5 x 4.184 x (26.2-100)]. Note that the only error you made here is you used the specific heat of steam and not water.
Your second term is for the delta H vap. You can't use 40.7 kJ/mol. That's what you're trying to find. It is (5 x heat vap)
Add these three terms and the sum = 0. I get about 2280 J/g