Calculate the quantity of energy required to change 3.00 mol of liquid water to steam at 100°C. The molar heat of vaporization of water is 40.6 kj/mol
energy=moles*heatVaporization.
To calculate the quantity of energy required to change 3.00 mol of liquid water to steam at 100°C, we can use the formula:
Energy required = Number of moles * Molar heat of vaporization
Plugging in the values:
Energy required = 3.00 mol * 40.6 kJ/mol
Energy required = 121.8 kJ
Therefore, the quantity of energy required would be 121.8 kJ.
To calculate the quantity of energy required to change liquid water to steam, we need to consider two steps: heating the water and then vaporizing it.
Step 1: Heating the water
The specific heat capacity of water is approximately 4.18 J/g°C. However, we are given the quantity of water in moles, so we need to convert it to grams. The molar mass of water (H2O) is approximately 18.02 g/mol.
3.00 mol of water = 3.00 mol × 18.02 g/mol = 54.06 g
To raise the temperature of liquid water from its initial temperature (let's assume it's 25°C) to its boiling point (100°C), we need to calculate the amount of heat using the formula:
q = m × c × ΔT
where:
q = amount of heat (in Joules)
m = mass of water (in grams)
c = specific heat capacity of water (in J/g°C)
ΔT = change in temperature (final temperature - initial temperature)
ΔT = 100°C - 25°C = 75°C
q = 54.06 g × 4.18 J/g°C × 75°C = 17,950.215 J
Step 2: Vaporizing the water
The molar heat of vaporization of water is given as 40.6 kJ/mol. We need to convert this to Joules:
40.6 kJ = 40.6 kJ × 1000 J/kJ = 40,600 J/mol
Now, we can calculate the amount of energy required to vaporize the water from liquid to steam:
q = n × ΔHvap
where:
q = amount of heat (in Joules)
n = number of moles of water
ΔHvap = molar heat of vaporization of water (in J/mol)
q = 3.00 mol × 40,600 J/mol = 121,800 J
Therefore, the total energy required to change 3.00 mol of liquid water to steam at 100°C is the sum of the energy required for heating and vaporization:
Total energy = heating energy + vaporization energy
Total energy = 17,950.215 J + 121,800 J = 139,750.215 J (rounded to four significant figures)
Hence, the quantity of energy required is approximately 139,750 J.