Mojtaba S. answered • 04/10/23
PhD in Engineering with 6+ Years of Teaching Experience for MCAT
To calculate the entropy change when water boils, we can use the following formula:
ΔS = q/T
where ΔS is the entropy change, q is the heat transferred, and T is the temperature in Kelvin.
First, we need to calculate the amount of heat transferred, which we can do using the enthalpy of vaporization and the number of moles of water that vaporize:
q = ΔHvap * n
where ΔHvap is the enthalpy of vaporization, and n is the number of moles of water that vaporize.
To find n, we can use the following equation:
n = m/M
where m is the mass of water and M is the molar mass of water.
n = 36.0 g / 18.0 g/mol = 2.00 mol
Now we can calculate the heat transferred:
q = ΔHvap * n = 40660 J/mol * 2.00 mol = 81320 J
Next, we need to convert the boiling temperature of water from Celsius to Kelvin:
T = 93°C + 273.15 = 366.15 K
Finally, we can calculate the entropy change:
ΔS = q/T = 81320 J / 366.15 K = 222.12 J/K
Therefore, the entropy change when 36.0g of water boils at 93°C is 222.12 J/K.
