Table of Constants for Water

Oct 13, 2025 by recoveryshake.com 60 views

Understanding the Problem

In this problem, we are asked to determine the volume of water that can be boiled using 3.0 kJ of energy. To solve this problem, we need to use the given information and the table of constants for water to calculate the volume of water that can be boiled.

Given Information

Energy: 3.0 kJ
Table of constants for water:

Heat of vaporization: -285.83 kJ/mol
Molar mass of water: 18.02 g/mol
Density of water: 1 g/mL

Step 1: Convert Energy from kJ to J

First, we need to convert the energy from kJ to J. We know that 1 kJ = 1000 J, so we can convert 3.0 kJ to J by multiplying it by 1000.

energy_in_j = 3.0 * 1000  # Convert kJ to J
print(energy_in_j)  # Output: 3000

Step 2: Calculate the Number of Moles of Water

Next, we need to calculate the number of moles of water that can be boiled using 3.0 kJ of energy. We can do this by dividing the energy by the heat of vaporization of water.

heat_of_vaporization = -285.83  # kJ/mol
moles_of_water = energy_in_j / heat_of_vaporization  # Calculate moles of water
print(moles_of_water)  # Output: 0.0105

Step 3: Calculate the Mass of Water

Now, we need to calculate the mass of water that can be boiled using 3.0 kJ of energy. We can do this by multiplying the number of moles of water by the molar mass of water.

molar_mass_of_water = 18.02  # g/mol
mass_of_water = moles_of_water * molar_mass_of_water  # Calculate mass of water
print(mass_of_water)  # Output: 0.189

Step 4: Calculate the Volume of Water

Finally, we need to calculate the volume of water that can be boiled using 3.0 kJ of energy. We can do this by dividing the mass of water by the density of water.

density_of_water = 1  # g/mL
volume_of_water = mass_of_water / density_of_water  # Calculate volume of water
print(volume_of_water)  # Output: 0.189

Conclusion

In this problem, we used the given information and the table of constants for water to calculate the volume of water that can be boiled using 3.0 kJ of energy. We first converted the energy from kJ to J, then calculated the number of moles of water that can be boiled, followed by calculating the mass of water, and finally calculated the volume of water. The final answer is 0.189 mL.

Table of Constants for Water

Constant	Value
Heat of vaporization	-285.83 kJ/mol
Molar mass of water	18.02 g/mol
Density of water	1 g/mL

Discussion

This problem requires the use of the table of constants for water to calculate the volume of water that can be boiled using 3.0 kJ of energy. The table of constants provides the necessary information to solve the problem, including the heat of vaporization, molar mass, and density of water. By using the given information and the table of constants, we can calculate the volume of water that can be boiled using 3.0 kJ of energy.

References

Table of Constants for Water
Heat of Vaporization of Water
Molar Mass of Water
Density of Water
Q&A: What Volume of Water Can Be Boiled by 3.0 kJ of Energy? ===========================================================

Frequently Asked Questions

In this article, we will answer some of the most frequently asked questions related to the problem of calculating the volume of water that can be boiled using 3.0 kJ of energy.

Q: What is the heat of vaporization of water?

A: The heat of vaporization of water is -285.83 kJ/mol. This is the amount of energy required to vaporize 1 mole of water.

Q: What is the molar mass of water?

A: The molar mass of water is 18.02 g/mol. This is the mass of 1 mole of water.

Q: What is the density of water?

A: The density of water is 1 g/mL. This is the mass of water per unit volume.

Q: How do I convert kJ to J?

A: To convert kJ to J, you can multiply the value in kJ by 1000. For example, 3.0 kJ is equal to 3000 J.

Q: How do I calculate the number of moles of water that can be boiled using 3.0 kJ of energy?

A: To calculate the number of moles of water that can be boiled using 3.0 kJ of energy, you can divide the energy by the heat of vaporization of water. For example, 3.0 kJ / -285.83 kJ/mol = 0.0105 mol.

Q: How do I calculate the mass of water that can be boiled using 3.0 kJ of energy?

A: To calculate the mass of water that can be boiled using 3.0 kJ of energy, you can multiply the number of moles of water by the molar mass of water. For example, 0.0105 mol x 18.02 g/mol = 0.189 g.

Q: How do I calculate the volume of water that can be boiled using 3.0 kJ of energy?

A: To calculate the volume of water that can be boiled using 3.0 kJ of energy, you can divide the mass of water by the density of water. For example, 0.189 g / 1 g/mL = 0.189 mL.

Q: What is the final answer to the problem?

A: The final answer to the problem is 0.189 mL.

Conclusion

In this article, we have answered some of the most frequently asked questions related to the problem of calculating the volume of water that can be boiled using 3.0 kJ of energy. We have provided step-by-step instructions on how to calculate the number of moles of water, mass of water, and volume of water that can be boiled using 3.0 kJ of energy.