Old fashion steam locomotives operated by boiling water and using the steam to drive a piston that moved the train. How much energy ( in MJ) is required to turn 824 gallons of water at 100 C into steam? The heat of fusion for water is 6.01 kJ/mol and the heat of vaporization is 40.7 kJ/mol. The units are MJ.
824 gallons x approx 3.8 L/gallon = approx 3000 L (but you need to use a better factor and not estimate as I've done.
Then 3000 L has a mass of about 3000 g.
q = mass H2O x heat vaporiation = ?
Note: If you use 40.7 kJ/mol you must use mass H2O in mol(which means convert from grams to mols), and the answer will come out in kJ. Convert to MJ.
To calculate the energy required to turn water into steam, we need to consider both the energy needed to heat the water and the energy needed to vaporize it.
First, we need to convert the volume of water from gallons to liters. Since 1 gallon is approximately equal to 3.78541 liters, we multiply 824 gallons by 3.78541 to get 3114.96 liters.
Next, we need to convert the mass of water from liters to moles. The molar mass of water (H2O) is approximately 18 grams per mole. We can multiply the volume of water in liters by the density of water (approximately 1 gram per milliliter) and divide that by the molar mass of water to get the number of moles of water. In this case, we get:
3114.96 liters * 1 gram/mL / 1000 grams/mole / 18 grams/mole = 171.94 moles of water
Now, let's calculate the energy required for each step:
1. Heating the water to its boiling point:
To heat water from 0°C (assuming that's the initial temperature) to 100°C, we need to use the heat capacity formula:
Q = m * ΔT * Cp
where Q is the energy, m is the mass, ΔT is the change in temperature, and Cp is the specific heat capacity of water, which is approximately 4.184 J/(g·°C).
Converting the mass from moles to grams, we have:
171.94 moles * 18 grams/mole = 3094.92 grams
ΔT = 100°C - 0°C = 100°C
Plugging in these values, we get:
Q1 = 3094.92 grams * 100°C * 4.184 J/(g·°C) = 1,293,070.752 J
Converting to megajoules (MJ), we divide by 1,000,000:
Q1 = 1,293,070.752 J / 1,000,000 = 1.29307 MJ
2. Vaporizing the water:
To vaporize the water, we need to use the enthalpy of vaporization, which is given as 40.7 kJ/mol.
The energy required to vaporize the water is given by:
Q2 = n * ΔHvap
where n is the number of moles and ΔHvap is the enthalpy of vaporization.
Plugging in the values, we have:
Q2 = 171.94 moles * 40.7 kJ/mol = 6,988.318 kJ
Converting to megajoules (MJ), we divide by 1,000:
Q2 = 6,988.318 kJ / 1,000 = 6.988318 MJ
Finally, we can add the two energies together to get the total energy required:
Total energy = Q1 + Q2 = 1.29307 MJ + 6.988318 MJ = 8.281388 MJ
Therefore, approximately 8.281388 MJ of energy is required to turn 824 gallons of water at 100°C into steam.