How much energy (heat) is required to convert 248 g of water from 0oC to 154oC? Assume that the water begins as a liquid, that the specific heat of water is 4.184 J/g.oC over the entire liquid range, that the specific heat of steam is 1.99 J/g.oC, and the heat of vaporization of water is 40.79 kJ/mol.
692kJ
To calculate the total energy required to convert 248 g of water from 0°C to 154°C, we need to consider the following steps:
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.
To calculate the amount of energy required to convert 248 g of water from 0oC to 154oC, we need to consider two steps:
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.
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.