Calculate the heat released when 73.5 g of steam at 122.0 C is converted to water at 39.0C. Assume that the specific heat of water is 4.184 J/g C, the specific heat of steam is 1.99 J/g C, and change in heat vap =40.79 kJ/mol for water.
q1 = heat released when steam @ 122 moves to 100.
q1 = mass steam x sp h steam x (Tfinal-Tinital)
q2 = heat released when steam condenses.
q2 = mass steam x heat vap
q3 = heat released when H2O @ 100 moves to 39 C.
q3 = mass H2O x sp h H2O x (Tfinal-Tinitial)
Then add q1 + q2 + q3.
To calculate the heat released when converting steam to water, we need to consider two separate processes: first, cooling the steam from its initial temperature to 100°C, and then condensing the steam at 100°C to water at the final temperature.
Let's break down the calculation into two parts:
1. Cooling the steam from 122.0°C to 100°C:
The heat released during this process is given by the equation:
Q1 = mass of steam * specific heat of steam * (final temperature - initial temperature)
Q1 = 73.5 g * 1.99 J/g°C * (100°C - 122.0°C)
Q1 = -3559.59 J (notice the negative sign as heat is being released)
2. Condensing the steam at 100°C to water at 39.0°C:
The heat released during this process is given by the equation:
Q2 = heat of vaporization * moles of water vapor
First, we need to find the moles of water vapor:
Moles of water vapor = mass of steam / molar mass of water
The molar mass of water is H2O, which is approximately 18.015 g/mol.
Moles of water vapor = 73.5 g / 18.015 g/mol ≈ 4.08 mol
Now, we can calculate Q2:
Q2 = 40.79 kJ/mol * 4.08 mol
= 166.6432 kJ = 166643.2 J
(Note: We converted kJ to J for consistency with the previous calculation)
Now, to find the total heat released, we add the heat released during both processes:
Total heat released = Q1 + Q2
= -3559.59 J + 166643.2 J
= 163083.61 J
Therefore, the heat released when converting 73.5 g of steam at 122.0°C to water at 39.0°C is approximately 163,084 J.