Assume that the pressure is one atmosphere and determine the heat in joules required to produce 3.23 kg of water vapor at 100.0 °C, starting with (a)3.23 kg of water at 100.0 °C and (b)3.23 kg of liquid water at 0.0 °C
a) q1 = heat needed to transform 100 C liquid water to 100 C steam.
q1 = mass water in grams x heat fusion
b)
q2 = heat needed to transform zero C liquid H2O to liquid water at 100 C.
q2 = mass H2O in grams x specific heat water x delta T (delta T is 100-0=100).
q3 = heat needed to vaporize the water
to steam.
q3 = mass H2O in grams x heat fusion.
For part b. total q = q2 + q3.
q2 = mass water x heat fusion.
To calculate the heat required to produce water vapor, we need to consider the phase changes involved. The specific heat capacity of water and the heat of vaporization are necessary for these calculations. The specific heat capacity of water is 4.18 J/g°C and the heat of vaporization is 40.7 kJ/mol.
(a) To determine the heat required to produce 3.23 kg of water vapor at 100.0 °C, starting with 3.23 kg of water at 100.0 °C, we can break down the process into two steps:
Step 1: Heating the water from 100.0 °C to its boiling point at 100 °C.
Heat required = mass × specific heat capacity × temperature change
Heat required = 3.23 kg × 4.18 J/g°C × (100.0 °C - 100.0 °C)
Heat required = 0 J
Step 2: Vaporizing the water at its boiling point.
Heat required = mass × heat of vaporization
Heat required = 3.23 kg × (40.7 kJ/mol / molar mass of water)
To find the molar mass of water, we use the molecular weight of hydrogen (1 g/mol) and oxygen (16 g/mol):
Molar mass of water = 2 × (1 g/mol) + 16 g/mol
Molar mass of water = 18 g/mol
Heat required = 3.23 kg × (40.7 kJ/mol / 18 g/mol)
Heat required = 7,212.85 kJ
To convert to joules:
Heat required = 7,212.85 kJ × 1,000 J/kJ
Heat required = 7,212,850 J
Therefore, the heat required to produce 3.23 kg of water vapor at 100.0 °C, starting with 3.23 kg of water at 100.0 °C, is 7,212,850 J.
(b) To determine the heat required to produce 3.23 kg of water vapor at 100.0 °C, starting with 3.23 kg of liquid water at 0.0 °C, we need to consider the process of raising the temperature to the boiling point at 100 °C and then vaporization.
Step 1: Heating the water from 0.0 °C to its boiling point at 100 °C.
Heat required = mass × specific heat capacity × temperature change
Heat required = 3.23 kg × 4.18 J/g°C × (100.0 °C - 0.0 °C)
Heat required = 1,354.678 J
Step 2: Vaporizing the water at its boiling point.
Heat required = mass × heat of vaporization
Heat required = 3.23 kg × (40.7 kJ/mol / 18 g/mol)
Heat required = 7,212.85 kJ
To convert to joules:
Heat required = 7,212.85 kJ × 1,000 J/kJ
Heat required = 7,212,850 J
Therefore, the heat required to produce 3.23 kg of water vapor at 100.0 °C, starting with 3.23 kg of liquid water at 0.0 °C is 7,214,204.678 J.
To determine the heat required to produce water vapor, we need to consider the phase changes and calculate the energy needed for each step. We'll break down the process into two main steps: heating the water to its boiling point and then turning it into vapor.
a) Heating 3.23 kg of water from 100.0 °C to its boiling point:
First, we need to calculate the heat required to raise the temperature of the water from 100.0 °C to its boiling point, which is 100.0 °C.
The specific heat capacity of water is approximately 4.18 J/g°C.
So, the heat required is given by the formula:
Heat = mass × specific heat capacity × temperature change.
Converting the mass of water to grams:
mass of water = 3.23 kg × 1000 g/kg = 3230 g
Temperature change:
ΔT = final temperature - initial temperature
ΔT = 100.0 °C - 100.0 °C = 0 °C
Heat = 3230 g × 4.18 J/g°C × 0 °C = 0 J (no heat is required)
b) Turning 3.23 kg of liquid water at 0.0 °C into vapor at 100.0 °C:
Now we need to calculate the heat required to convert liquid water at its boiling point (100.0 °C) into vapor at the same temperature.
The heat required for this phase change is given by the formula:
Heat = mass × heat of vaporization.
The heat of vaporization for water is approximately 2260 J/g.
Converting the mass of water to grams:
mass of water = 3.23 kg × 1000 g/kg = 3230 g
Heat = 3230 g × 2260 J/g = 7,304,980 J or 7.305 MJ (to three significant figures).
Therefore, starting with 3.23 kg of liquid water at 0.0 °C, it would require approximately 7.305 megajoules of energy to produce 3.23 kg of water vapor at 100.0 °C.