Question
Find the increase in entropy of 1.00 kg of liquid nitrogen that starts at its boiling temperature, boils, and warms to at constant pressure.
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Solution video
OpenStax College Physics, Chapter 15, Problem 54 (Problems & Exercises)
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This is College Physics Answers with Shaun Dychko. One kilogram of liquid nitrogen initially at its boiling point of negative 195.8 degrees Celsius is going to boil and then increases its temperature in the gaseous state at constant pressure to a temperature of 20.0 degrees Celsius. So the temperatures are converted into Kelvin by adding 273.15 and we need to know what the latent heat of vaporization is for nitrogen—201 times 10 to the 3 joules per kilogram— and the specific heat is 1040 joules per kilogram per Celsius degree. So we are gonna find the total change in entropy and so that's gonna be the sum of the change in entropy due to the boiling or vaporization plus the change in entropy due to the increase in temperature from the boiling point T 1 to the final temperature of T 2. So the change in entropy due to boiling is going to be the heat that it absorbs divided by its temperature and the heat it absorbs will be its mass times the latent heat of vaporization and all that gets divided by temperature T 1 that doesn't change while it's boiling. And then the change in entropy due to its increase in temperature after it changes phase into the gaseous state is the total heat absorbed which will be its mass times its specific heat times the change in temperature, T 2 minus T 1, all divided by the average temperature. So that's mc times T 2 minus T 1 over the sum of the temperatures divided by 2 and that's the same as saying 2 divided by that sum of temperatures. So we have 2mc times T 2 minus T 1 all over T 1 plus T 2 is the change in temperature due to the increase or sorry... change in entropy due to the increase in temperature. So the final change in entropy then is mL v over T 1 plus 2mc times T 2 minus T 1 all over T 1 plus T 2; I factored out the m and then plugged in numbers. So that's 1.00 kilogram times 201 times 10 to the 3 joules per kilogram of latent heat of vaporization divided by 77.35 Kelvin— which is the boiling point of Nitrogen— plus 2 times 1040 joules per kilogram per Celsius degree times the change in temperature divided by the sum of the temperatures and we get 3.81 times 10 to the 3 joules per Kelvin is the total change in entropy.