calculate the amount of energy (in kj) necessary to convert 457 g of liquid water from 0 Celecius to water vapor at 167 celcius. the molar heat of vaporization (Hvap) of water is 40.79 jk/mol. the specific heat for water is 4.187 j/g celcius and for steam 1.99 j/mol celcius. (assume that the specific heat values do not change over the range of temperature in the problem)
Note the correct spelling of celsius.
There are two equations you need to do this.
1. WITHIN a phase, q = mass x specific heat x (Tfinal-Tinitial)
2a. At the phase change, q = mass x heat fusion at melting point or
2b. At the boiling point, q = mass x heat vaporization.
For example, within a phase would be from zero C t 100 C it is liquid water.
q = mass x specific heat liquid water x (Tf-Ti). Tf = 100; Ti = 0.
So q1 = zero C to 100 C.
q2 = convert liquid water to steam at 100 C.
q3 = steam from 100 C to final T.
Then add q1 + q2 + q3.
if the length of one side of a square is 12.0, what is the perimeter of the square?
To calculate the amount of energy required to convert liquid water to water vapor, we need to consider two steps: raising the temperature of liquid water from 0 °C to its boiling point, and then vaporizing it.
Step 1: Heating the liquid water from 0 °C to its boiling point.
To find the energy required to raise the temperature, we can use the formula:
Q = m * C * ΔT
where:
Q is the energy in joules (J)
m is the mass of water in grams (457 g)
C is the specific heat capacity of water in J/(g °C) (4.187 J/g °C)
ΔT is the change in temperature in °C (Boiling point - Initial temperature)
ΔT = 100 °C - 0 °C = 100 °C
Substituting the values into the formula:
Q1 = (457 g) * (4.187 J/g °C) * (100 °C)
Converting the result to kJ:
Q1 = (457 g) * (4.187 J/g °C) * (100 °C) / 1000
Q1 = 1913.679 kJ
Step 2: Vaporizing the liquid water at its boiling point.
To find the energy required for vaporization, we can use the formula:
Q = m * Hvap
where:
Q is the energy in joules (J)
m is the mass of water in grams (457 g)
Hvap is the molar heat of vaporization of water in J/mol (40.79 J/mol)
First, we need to calculate the number of moles of water:
moles = mass / molar mass
molar mass of water (H2O) = 18.015 g/mol
moles = 457 g / 18.015 g/mol
moles ≈ 25.41 mol
Substituting the values into the formula:
Q2 = (457 g) * (40.79 J/mol)
Converting the result to kJ:
Q2 = (457 g) * (40.79 J/mol) / 1000
Q2 = 18.650 kJ
Finally, to find the total energy required, we sum up Q1 and Q2:
Total Energy = Q1 + Q2
Total Energy = 1913.679 kJ + 18.650 kJ
Total Energy ≈ 1932.329 kJ
Therefore, the amount of energy necessary to convert 457 g of liquid water from 0 °C to water vapor at 167 °C is approximately 1932.329 kJ.
To calculate the amount of energy required to convert liquid water to water vapor, we need to consider two steps:
1. Heating the liquid water from 0°C to 100°C
2. Vaporizing the water at 100°C to water vapor at 167°C
Step 1: Calculate the energy required to heat the liquid water from 0°C to 100°C.
The specific heat capacity of water is given as 4.187 J/g°C.
The mass of water is 457 g.
The temperature change is 100°C - 0°C = 100°C.
Using the formula: Q = m * C * ΔT (where Q is the heat energy, m is the mass, C is the specific heat capacity, and ΔT is the temperature change)
Q1 = 457 g * 4.187 J/g°C * 100°C
Q1 = 190550 Joules
To convert the energy from Joules to kJ, divide by 1000.
Q1 = 190.55 kJ
Step 2: Calculate the energy required to vaporize the water at 100°C to water vapor at 167°C.
The molar heat of vaporization (Hvap) of water is given as 40.79 kJ/mol.
We first need to calculate the number of moles of water.
Using the molar mass of water (H₂O) which is approximately 18.015 g/mol:
Moles of water = Mass of water / Molar mass of water
Moles of water = 457 g / 18.015 g/mol
Moles of water ≈ 25.42 moles
The temperature change is 167°C - 100°C = 67°C.
Using the formula: Q = n * Hvap * ΔT (where Q is the heat energy, n is the number of moles, Hvap is the molar heat of vaporization, and ΔT is the temperature change)
Q2 = 25.42 moles * 40.79 kJ/mol * 67°C
Q2 ≈ 69,770.722 kJ
The total energy required is the sum of Q1 and Q2:
Total energy = Q1 + Q2
Total energy = 190.55 kJ + 69,770.722 kJ
Total energy ≈ 69,961.272 kJ
Therefore, approximately 69,961.272 kJ of energy is necessary to convert 457 g of liquid water from 0°C to water vapor at 167°C.