In an experiment, 230 kJ of heat energy are released to the surroundings when steam is converted to liquid water. Calculate the number of grams of water condensed. The following data may be of use:
vap = 2260 J/g; Hfus = 334 J/g; Cp(ice) = 2.06 J/g • °C; Cp(water) = 4.18 J/g • °C;
Cp(steam) = 1.99 J/g • °C
Same song. Third verse.
To calculate the number of grams of water condensed, we need to consider the different heat changes that occur during the process of steam being converted to liquid water.
First, we need to calculate the amount of heat released during the condensation process. Since 230 kJ of heat energy are released, we convert this value to J:
230 kJ = 230,000 J
Next, we need to determine the amount of heat released during the vaporization process. We know that the heat of vaporization (ΔHvap) is 2260 J/g. Therefore, the heat released during vaporization can be calculated using the equation:
Qvap = m × ΔHvap
where Qvap is the heat released during vaporization, m is the mass of the water vapor, and ΔHvap is the heat of vaporization.
Next, we need to determine the amount of heat released during the fusion process. We know that the heat of fusion (ΔHfus) is 334 J/g. Therefore, the heat released during fusion can be calculated using the equation:
Qfus = m × ΔHfus
where Qfus is the heat released during fusion, m is the mass of the water condensed, and ΔHfus is the heat of fusion.
Lastly, we need to calculate the amount of heat released during the cooling process. The specific heat capacity (Cp) is a property that represents the amount of heat required to raise the temperature of a substance by 1°C. We know the specific heat capacities of ice, water, and steam are 2.06 J/g • °C, 4.18 J/g • °C, and 1.99 J/g • °C, respectively.
To calculate the heat released during cooling, we need to consider the temperature change from the initial temperature of the steam to the final temperature of the liquid water. Assuming the initial temperature of the steam is 100°C (boiling point) and the final temperature of the liquid water is 0°C (fusion point), we can calculate the heat released as follows:
Qcool = m × Cp(steam) × (100 - 0) + m × Cp(water) × (0 - 100) + m × Cp(ice) × (0 - (-100))
where Qcool is the heat released during cooling, m is the mass of the water condensed, Cp(steam), Cp(water), and Cp(ice) are the specific heat capacities of steam, water, and ice, respectively.
Now, we can rewrite the total heat released equation as:
230,000 J = Qvap + Qfus + Qcool
Finally, rearranging the equation to solve for the mass of water condensed (m), we get:
m = (230,000 J − Qcool − Qvap) / (ΔHfus)
Substituting the values we have for the specific heat capacities, heat of vaporization, and heat of fusion, you can now calculate the number of grams of water condensed.