A freezer sits in a large room. A container of liquid water,...

A freezer sits in a large room. A container of liquid water, with a volume of 2.0 L and at a temperature of 0°C, is placed in the freezer, where it is completely converted into ice. Throughout the freezing process, the interior of the freezer is maintained at a temperature of 0°C. (a) Find the change in entropy (in kJ/K) of the water during the freezing process. (b) Find the change in entropy (in kJ/K) of the interior of the freezer (not including the water) during the freezing process. (c) The freezer's coefficient of performance (cooling mode) is 4.00. How much heat (in J) does the freezer expel to the room in which the freezer sits during the freezing process? Give the absolute value. (d) If the room that houses the freezer stays at a constant temperature of 20.0°C, what is the total entropy change (of the water, freezer interior, and the room combined, in kJ/K) during the freezing process?

Transcript

00:01 In this example, i'm going to be looking at the relationship between heat, temperature, and entropy.

00:07 Okay, so what we have is our system is we have a freezer in a room.

00:12 Okay, this freezer is kept at a constant temperature of zero degrees celsius.

00:18 Okay, and into this freezer we are going to place a volume of water, which is also initially at zero degrees celsius.

00:28 We're going to allow it to cool until it becomes completely ice in the freezer.

00:34 Okay, and what we want to do is find the change in entropy of both the water slash ice and the freezer during this process.

00:42 Okay, then we're given the coefficient of performance of the refrigerator and we're given the temperature of the room.

00:49 Okay, then we want to find the heat, total heat expelled into the room, and then the change in entropy of the room, and then finally we're going to find the total change in entropy of the whole system.

00:58 So, that's the room.

01:01 Okay, the freezer, and then our little volume of water will be number three.

01:07 So, we have water in there.

01:09 Okay, so the initial temperature of the freezer equals the temperature of the water equals zero degrees celsius.

01:21 Our volume of water is 2 .0 liters.

01:27 Okay, the heat of fusion for water or ice is 333 kilojoules per kilogram.

01:39 Okay, and the coefficient of performance of the refrigerator is 4 .00.

01:46 I don't know why they give that to you with three significant figures when we only have two with the volume, but they do.

01:53 Okay, so how do we approach this problem? first, we need to find the heat that is lost by the water as it is transformed into ice.

02:01 That'll be equal to the heat gained by the freezer as it does work to freeze the water into ice.

02:08 Then, we're going to use the coefficient of performance to determine the total work we had to put into the freezer to perform this transformation from water to ice.

02:23 Then, we're finally going to find the heat expelled to the room and then find the change in entropy of the room.

02:30 We know that the temperature of the room, tr, equals 20 degrees celsius.

02:37 Right, we're going to be working with kelvin, so i'll just convert your temperatures now.

02:41 Okay, t freezer equals t water equals zero c equals 273 .15 k.

02:51 Okay, the temperature of the room in kelvin is 293 .15 k.

02:58 Okay, so the heat lost by the water, that's going to be the mass of the water times the latent heat of fusion.

03:05 We don't have mass, but we have volume and we know the density of the water.

03:09 At zero degrees, the density is something like 998 grams per liter.

03:18 I'm just going to use one kilogram with rounding.

03:20 It's not going to affect the problem anyway.

03:22 Okay, so the heat lost by the water, q water, this will be part a, equals the mass of my water.

03:32 Okay, so one kilogram times two, um, two point, no, sorry, one kilogram per liter times two liters gives me two kilograms of water, two point oh, times my latent heat of fusion, lf, times 333 kilojoules per kilogram.

03:55 We have a loss of 666 kilojoules...

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