A pot contains 243 g of water at 78°C. If this water is heated and all evaporates to form steam at 100°C, what is the change in the entropy of the H2O?
To calculate the change in entropy of H2O, we need to use the equation:
ΔS = q/T
where ΔS is the change in entropy, q is the heat absorbed or released, and T is the temperature.
In this case, we have water at 78°C being heated until it completely evaporates into steam at 100°C. To find the change in entropy, we need to consider the heat absorbed during this process.
First, we need to calculate the heat absorbed by the water to heat it up from 78°C to its boiling point at 100°C. We can use the heat capacity formula:
q1 = m * C * ΔT
where q1 is the heat absorbed, m is the mass of the water, C is the specific heat capacity of water, and ΔT is the change in temperature.
The specific heat capacity of water is approximately 4.18 J/g°C.
Calculating q1:
q1 = 243 g * 4.18 J/g°C * (100°C - 78°C)
q1 = 243 g * 4.18 J/g°C * 22°C
q1 ≈ 21,575.68 J
Next, we need to calculate the heat absorbed by the water to change its phase from liquid to gas. We can use the formula:
q2 = m * ΔHvap
where q2 is the heat absorbed, m is the mass of the water, and ΔHvap is the heat of vaporization of water.
The heat of vaporization of water is approximately 40.7 kJ/mol. To convert this to J/g, we need to divide it by the molar mass of water, which is approximately 18 g/mol.
Calculating q2:
q2 = 243 g * (40.7 kJ/mol / 18 g/mol)
q2 ≈ 540.55 kJ
Converting q2 from kJ to J:
q2 = 540.55 kJ * 1000 J/kJ
q2 ≈ 540,550 J
Now, we have the total heat absorbed during the process:
q_total = q1 + q2
q_total = 21,575.68 J + 540,550 J
q_total ≈ 561,125.68 J
Finally, we can calculate the change in entropy using the equation:
ΔS = q_total / T
where T is the final temperature of the system, which is 100°C.
Calculating ΔS:
ΔS = 561,125.68 J / (100°C + 273.15 K)
ΔS ≈ 1583.27 J/K
Therefore, the change in entropy of the water is approximately 1583.27 J/K.