A mass m = 0.068 kg of benzene vapor (Lv = 3.94x105 J/kg) at its boiling point of 80.1°C is to be condensed by mixing with water at 50.0°C. What is the minimum mass of water required to condense all of the benzene vapor? Assume the mixing and condensation take place is a perfectly insulating container.
mL= =m₁c ΔT
m₁ = mL/c ΔT =
=0.068•3.94•10⁵/4183•30.1=
=0.213 kg
To find the minimum mass of water required to condense all of the benzene vapor, we need to calculate the amount of heat that needs to be transferred from the benzene vapor to the water.
The heat transferred during the condensation process can be calculated using the formula:
Q = m * Lv
where:
Q is the amount of heat transferred (in Joules),
m is the mass of the substance being condensed (in kg),
Lv is the latent heat of vaporization (in J/kg).
In this case, the mass of the benzene vapor m is given as 0.068 kg, and the latent heat of vaporization Lv is given as 3.94x10^5 J/kg.
Let's calculate the amount of heat transferred:
Q = 0.068 kg * 3.94x10^5 J/kg
Q ≈ 26,792 J
Now, we need to find the minimum mass of water required to absorb this amount of heat.
The specific heat capacity of water is approximately 4.18 J/g°C.
The temperature change of the water is (80.1 - 50.0) = 30.1°C.
To find the mass of water, we can use the equation:
Q = m * c * ΔT
where:
m is the mass of water (in grams),
c is the specific heat capacity of water (in J/g°C),
ΔT is the temperature change of the water (in °C).
Let's calculate the mass of water:
26,792 J = m * 4.18 J/g°C * 30.1°C
m ≈ 205 grams
Therefore, the minimum mass of water required to condense all of the benzene vapor is approximately 205 grams.
To determine the minimum mass of water required to condense all of the benzene vapor, we need to calculate the heat transfer from the benzene vapor to the water and then use that information to find the mass of water required.
The heat transfer from the benzene vapor to the water can be calculated using the heat transfer equation:
Q = mcΔT
Where:
Q is the heat transfer (in Joules),
m is the mass of the substance (in kilograms),
c is the specific heat capacity of the substance (in J/kg°C), and
ΔT is the change in temperature (in °C).
First, we need to calculate the heat transfer from the benzene vapor to reach its boiling point of 80.1°C. We can use the equation:
Q1 = m1 × Lv
Where:
Q1 is the heat transfer required to raise the temperature of the benzene vapor to its boiling point,
m1 is the mass of the benzene vapor,
and Lv is the latent heat of vaporization of benzene (3.94x10^5 J/kg).
Q1 = 0.068 kg × (3.94x10^5 J/kg)
Q1 = 26,872 J
Now, we need to calculate the heat transfer from the benzene vapor at its boiling point to the water at 50.0°C to condense the vapor. We can use the equation:
Q2 = m2 × cw × ΔT
Where:
Q2 is the heat transfer required to condense the benzene vapor,
m2 is the mass of the water required,
cw is the specific heat capacity of water (4186 J/kg°C),
and ΔT is the temperature difference between the water and the boiling point of benzene (80.1°C - 50.0°C).
Q2 = m2 × 4186 J/kg°C × (80.1°C - 50.0°C)
To condense all of the benzene vapor, the total heat transfer (Q) should be equal to Q1 + Q2.
Q = Q1 + Q2
Q = 26,872 J + (m2 × 4186 J/kg°C × 30.1°C)
Since the system is perfectly insulated, all the heat transfer from the vapor to the water is utilized for condensation (Q = 0). Therefore, we can write:
0 = 26,872 J + (m2 × 4186 J/kg°C × 30.1°C)
Now, solve for m2:
m2 = -26,872 J / (4186 J/kg°C × 30.1°C)
m2 ≈ 0.202 kg
So, the minimum mass of water required to condense all of the benzene vapor is approximately 0.202 kg.