An ice cube at a temperature of 00 degrees celsius and a...

An ice cube at a temperature of 0.0 degrees Celsius and a mass of .060 kg is dropped into the ocean. The temperature of the ocean is 20 degrees Celsius. a) As soon as the ice cube is immersed in the warmer ocean water it begins to melt. Calculate the change in entropy of the ice after it has just finished undergoing the phase transition from solid to liquid, but its temperature is still 0.0 degrees Celsius. Use L_f = 334.000 J/kg for water. b) Using your answer to part a, calculate the change in entropy of the ocean water assuming the change in temperature due to the addition of one ice cube is negligible and therefore ocean temperature can be considered constant. (Be careful with positive and negative signs. If heat is added to a body Q is positive. If heat is removed from a body Q is negative). c) In part a, we assumed that the ice would melt due to heat transfer from the surrounding ocean. Now consider the reverse process. Energy would still be conserved if heat flowed in the opposite direction, from the ice cube into the ocean. Use your answers to parts a and b above to show that our assumption was correct and that the reverse direction of heat transfer is impossible according to the 2nd law of thermodynamics.

Transcript

00:01 So this question deals with the concept of entropy.

00:05 We're going to use a definition of entropy that relates to a heat transfer, which says that the change in entropy of an object or a system is equal to the heat that enters or leaves that system divided by the temperature of that system.

00:28 And there's a thing about entropy that we're going to refer to in part c, which is called the second law of thermodynamics that says that the change in entropy, you know, for the universe as a whole, or if you have an isolated system, for that isolated system, let's just say it's for the universe as a whole and we'll use that same formulation for an isolated system is going to be greater than or equal to zero.

00:55 That is the requirement of the second law of thermodynamics.

00:58 So let's see what's going on here for part a.

01:01 It says we're supposed to calculate the change in entropy for an ice cube.

01:07 Well, the ice cube is gaining entropy since heat transfer is positive.

01:13 And so the change in entropy for the ice cube is going to equal the heat going into it, which we have to calculate it, actually.

01:22 Let's go ahead and calculate that as an aside here.

01:24 The heat going into the ice cube is going to equal the 3 .34 times 10 to the 5th joules per kilogram times the number of kilograms, which is here given to us as 0 .060 kilograms.

02:00 And that works out to be 20 ,000 .040 jules.

02:10 By the way, i used the specific heat that i had to look it up.

02:13 Sorry, latent heat of fusion, i should say.

02:16 There was in this version of problem, it looks like it says 334 .0 joules per kilogram, but that is either a misprint or it's using a continental system where you use.

02:32 The decimal point there to, you know, for the thousands place.

02:38 I went ahead and just wrote it in terms of scientific notation here, so there would be no ambiguity.

02:47 Anyway, so now we need to calculate the change in entropy, and that's going to equal our heat, which is going into the ice cube, so that is a positive value of 20 ,040 joules, and then that divided by the temperature.

03:04 It's zero celsius, and so that works out to be the kelvin temperature of 273 .15 kelvin.

03:15 And so if you do that calculation, which you end up with is the change in entropy of the water is 73 .366.

03:31 Duels per kelvin.

03:36 So now we need to calculate the change in entropy of the surrounding water.

03:45 Those will say this is for the ice.

03:49 So just distinguish it.

03:50 And for the surrounding water, well, the magnitude of the heat is the same...

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