Make a table:
Heat
1) to heat water to 100C
2) to change water to steam at 100C
3) to heat steam from 100C to 154C
figure all three of those, and add.
I will be happy to critique your thinking.
Heat
1) to heat water to 100C
2) to change water to steam at 100C
3) to heat steam from 100C to 154C
figure all three of those, and add.
I will be happy to critique your thinking.
1. Heating the water from 0°C to its boiling point:
The specific heat of water is 4.184 J/g.°C.
The boiling point of water is 100°C.
Energy required = mass × specific heat × temperature change
= 248 g × 4.184 J/g.°C × (100°C - 0°C)
= 104,090.24 J
2. Converting the water from liquid to steam:
The heat of vaporization of water is 40.79 kJ/mol.
The molar mass of water is approximately 18.015 g/mol.
The boiling point of water is 100°C.
Number of moles of water = mass / molar mass
= 248 g / 18.015 g/mol
≈ 13.764 mol
Energy required = number of moles × heat of vaporization
= 13.764 mol × 40.79 kJ/mol
≈ 562.437 kJ
≈ 562,437 J
3. Heating the steam from its boiling point to 154°C:
The specific heat of steam is 1.99 J/g.°C.
Initial temperature = boiling point of water = 100°C
Final temperature = 154°C
Energy required = mass × specific heat × temperature change
= 248 g × 1.99 J/g.°C × (154°C - 100°C)
= 26,430.24 J
Total energy required = Energy for heating water + Energy for phase change + Energy for heating steam
= 104,090.24 J + 562,437 J + 26,430.24 J
≈ 692,957 J
Therefore, approximately 692,957 Joules of energy (heat) is required to convert 248 g of water from 0°C to 154°C.
Step 1: Heating the water from 0oC to 100oC
Step 2: Converting the water from 100oC to 154oC
Step 1: Heating the Water from 0oC to 100oC
To calculate the energy required to heat the water, we use the formula:
Q = m * c * ΔT
where:
Q is the heat energy (in joules)
m is the mass of the water (in grams)
c is the specific heat capacity of water (in J/g.oC)
ΔT is the change in temperature (in oC)
Substituting the given values:
m = 248 g
c = 4.184 J/g.oC
ΔT = (100oC - 0oC) = 100oC
Q1 = 248 g * 4.184 J/g.oC * 100oC
Q1 = 103,708.8 J
So, the energy required to heat the water from 0oC to 100oC is 103,708.8 J.
Step 2: Converting the Water from 100oC to 154oC
While going from 100oC to 154oC, the water undergoes a change in phase from liquid to vapor. To calculate the energy required for this phase change, we will use the formula:
Q = m * ΔHvap
where:
Q is the heat energy (in joules)
m is the mass of the water (in grams)
ΔHvap is the heat of vaporization of water (in J/g)
Since the heat of vaporization is given in kJ/mol, we need to convert it to J/g.
To convert from kJ/mol to J/g, we need to divide the given value by the molar mass of water.
The molar mass of water (H2O) is:
2(atomic mass of hydrogen) + atomic mass of oxygen
2(1.008 g/mol) + 16.00 g/mol = 18.016 g/mol
Now, converting the heat of vaporization:
40.79 kJ/mol * (1000 J/1 kJ) / (18.016 g/mol) = 2260.31 J/g
Substituting the values:
m = 248 g
ΔHvap = 2260.31 J/g
Q2 = 248 g * 2260.31 J/g
Q2 = 560,360.88 J
So, the energy required to convert the water from 100oC to 154oC is 560,360.88 J.
Total Energy Required:
To find the total energy, we sum up the energy required for both steps:
Total Energy = Q1 + Q2
Total Energy = 103,708.8 J + 560,360.88 J
Total Energy = 664,069.68 J
Therefore, the total energy required to convert 248 g of water from 0oC to 154oC is approximately 664,069.68 J.