What is the total energy transfer(in KJ) when 10.00 pounds of water vapor at 110 degrees C becomes ice at 253K? Assume that the specific gravity of water = 1.00. The answer I got was around 12,000 KJ. Is this correct?
10 pounds of water? Goodness.
10 lbs *1kg/2.2lbs=you do it.
Then,
Heatcoolingsteamto100C=masswater*specificheatsteam*(100-110)
heat condensing steam= masssteam*HvSteam
Heatreleasedcoolingwater= masswater*specificheatwater*(0-100)
all those should be negative, add them,that is the heat gained total, so the heat released willbe the negative of that.
To calculate the total energy transfer during the phase change from water vapor to ice, you need to consider two processes: heating and cooling.
First, you need to calculate the energy required to heat 10.00 pounds of water vapor from 110 degrees Celsius to its boiling point (100 degrees Celsius) using the specific heat capacity of water. The specific heat capacity of water is approximately 4.18 J/g°C or 1 cal/g°C.
Let's convert the pounds to grams:
10.00 pounds = 4535.92 grams (1 pound = 453.59 grams).
Now, calculate the energy required to heat the water vapor:
q1 = mass × specific heat capacity × temperature change
= 4535.92 g × 4.18 J/g°C × (100 - 110)°C
= -189,053.76 J
Next, you need to calculate the energy required to change the water vapor at its boiling point (100 degrees Celsius) to water at the same temperature. This is known as the heat of vaporization. For water, the heat of vaporization is approximately 40.7 kJ/mol or 2260 J/g.
Let's calculate the number of moles of water vapor:
moles = mass / molar mass
= 4535.92 g / 18.02 g/mol
≈ 251.67 mol
Now, calculate the energy required for the phase change:
q2 = moles × heat of vaporization
= 251.67 mol × 2260 J/g
= 567,860.2 J
Finally, you need to calculate the energy released when the water cools down from 100 degrees Celsius to 0 degrees Celsius and solidifies into ice. The specific heat capacity of water is used again in this process.
Calculate the energy required to cool the water:
q3 = mass × specific heat capacity × temperature change
= 4535.92 g × 4.18 J/g°C × (0 - 100)°C
= -1,898,244.96 J
Now, add up the energies from each step to find the total energy transfer during this phase change:
Total energy transfer = q1 + q2 + q3
= -189,053.76 J + 567,860.2 J + (-1,898,244.96 J)
= -1,519,438.52 J
≈ -1,519.44 kJ
So, the total energy transfer in KJ is -1,519.44 KJ. Note that the negative sign indicates an energy release or loss.
Therefore, it seems that the answer you got, around 12,000 KJ, is incorrect.
To calculate the total energy transfer when water vapor condenses into ice, we need to consider both the heat required to cool the water vapor from 110 degrees Celsius to 0 degrees Celsius and the heat required for the phase change from water vapor to liquid water. Then, we need to consider the heat required to freeze the liquid water into ice at 0 degrees Celsius.
Here's the step-by-step calculation:
1. Calculate the heat required to cool the water vapor from 110°C to 0°C:
- The specific heat capacity of water vapor is 1.996 J/g·°C.
- The initial temperature is 110°C, and the final temperature is 0°C.
- The mass of the water vapor is 10.00 pounds, which is approximately 4535.92 grams.
- The formula to calculate heat is Q = m * c * ΔT, where Q is the heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
- Plugging in the values, we get Q1 = 4535.92 g * 1.996 J/g·°C * (0°C - 110°C).
2. Calculate the heat required for the phase change from water vapor to liquid water:
- The heat of vaporization of water is 40.7 kJ/mol.
- The molar mass of water is approximately 18.015 g/mol.
- Using a conversion factor, we can calculate the mass of water in moles: n = m / M, where n is the number of moles, m is the mass, and M is the molar mass.
- Plugging in the values, we get n = 4535.92 g / 18.015 g/mol.
- Now, we can calculate the heat using Q = n * ΔHvap, where ΔHvap is the heat of vaporization.
- Plugging in the values, we get Q2 = (4535.92 g / 18.015 g/mol) * 40.7 kJ/mol.
3. Calculate the heat required to freeze the liquid water into ice at 0°C:
- The heat of fusion of water is 6.01 kJ/mol.
- We can use the same mass in moles calculated in step 2 to calculate the heat using Q = n * ΔHfus.
- Plugging in the values, we get Q3 = (4535.92 g / 18.015 g/mol) * 6.01 kJ/mol.
4. Calculate the total energy transfer:
- Add up the three calculated values: Qtotal = Q1 + Q2 + Q3.
Based on your answer of approximately 12,000 KJ, it seems you may have made an error in one or more of the calculations. To verify, it would be helpful to see your work for each calculation.