How long would it take for 1.50mols of water at 100.0deg C to be converted completely into steam if heat were added at a constant rate of 16.0j/s
How much heat is required? Convert 1.50 moles to grams if heat of vaporization you use is in J/grams or leave it in mols if heat of vaporization you use is in J/mole.
q = mass x heat of vaporization.
Then q x (1 sec/16.0 J) = ?? sec.
Check me out on a little over an hour?
Spec heat of ice: 2.09 j/(g*deg C)
Spec heat of liq water: 4.18J/(g*deg C)
Enthalpy of fusion: 334 J/g
Enthalpy of vaporization: 2250 J/g
Answer in Minutes = 63 minutes. Thanks
The exact answer rounded to the correct digits is 63.3 minutes =]
What is the vapor pressure of benzene at 23◦C (about room tempera- ture)? The normal boiling point of benzene is 80◦C and its molar heat of vaporization is 30.8 kJ/mol.
To determine the time it would take for 1.50 moles of water to be completely converted into steam, we need to use the specific heat capacity and latent heat of vaporization.
First, let's calculate the energy required to convert the water into steam. We know the heat capacity of water (C) is 4.18 J/g°C, and the molar mass of water is approximately 18 g/mol. So, the molar heat capacity of water (Cm) can be calculated as follows:
Cm = C × molar mass
Cm = 4.18 J/g°C × 18 g/mol
Cm = 75.24 J/mol°C
Next, we need to calculate the energy required to raise the temperature of 1.50 moles of water from 100.0°C to its boiling point. The temperature change (ΔT) is the final temperature (boiling point) minus the initial temperature:
ΔT = boiling point - initial temperature
ΔT = 100°C - 0°C
ΔT = 100°C
To calculate the energy (Q1) required to raise the temperature of the water, we use the formula:
Q1 = Cm × moles × ΔT
Q1 = 75.24 J/mol°C × 1.50 mol × 100°C
Q1 = 11286 J
Now, let's calculate the energy required for the phase change from liquid to gas (steam). The latent heat of vaporization (L) for water is approximately 40.7 kJ/mol, which is equivalent to 40,700 J/mol.
The energy (Q2) required for the phase change from liquid to gas is calculated by multiplying the moles of water (n) with the latent heat of vaporization:
Q2 = n × L
Q2 = 1.50 mol × 40,700 J/mol
Q2 = 61,050 J
Finally, we can calculate the total time required (t) to convert the water into steam by dividing the total energy required (Q1 + Q2) by the heat added per second (16.0 J/s):
t = (Q1 + Q2) / heat added per second
t = (11286 J + 61050 J) / 16.0 J/s
t ≈ 7233 s
So, it would take approximately 7233 seconds (or about 2 hours) for 1.50 moles of water at 100.0°C to be completely converted into steam if heat were added at a constant rate of 16.0 J/s.