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 45.0°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.
To find the minimum mass of water required to condense all of the benzene vapor, we need to calculate the heat released by the benzene vapor as it condenses to liquid state.
The heat released during condensation can be calculated using the formula:
Q = m * Lv
where Q is the heat released, m is the mass of benzene vapor, and Lv is the latent heat of vaporization of benzene.
Given:
m = 0.068 kg (mass of benzene vapor)
Lv = 3.94x10^5 J/kg (latent heat of vaporization of benzene)
Substituting these values into the formula, we can calculate the heat released:
Q = 0.068 kg * 3.94x10^5 J/kg
Q = 26792 J
Now, we also need to calculate the heat absorbed by the water to raise its temperature from 45.0°C to the boiling temperature of benzene, which is 80.1°C.
The heat absorbed or gained by the water can be calculated using the formula:
Q = m * c * ΔT
where Q is the heat absorbed, m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature.
Given:
ΔT = 80.1°C - 45.0°C = 35.1°C
c = 4186 J/kg°C (specific heat capacity of water)
Substituting the values into the formula, we can calculate the heat absorbed by the water:
Q = m * c * ΔT
We can rearrange the formula to solve for m:
m = Q / (c * ΔT)
m = 26792 J / (4186 J/kg°C * 35.1°C)
Calculating the value:
m ≈ 0.183 kg
Therefore, the minimum mass of water required to condense all of the benzene vapor is approximately 0.183 kg.