WEBVTT 00:00:00.087 --> 00:00:02.850 This is College Physics Answers with Shaun Dychko 00:00:03.410 --> 00:00:06.436 An aluminum bowl with some soup is placed in a freezer, 00:00:06.705 --> 00:00:10.581 and the soup and bowl have an initial temperature of 25 degrees Celsius. 00:00:11.112 --> 00:00:16.247 And we’re told that 377 kiloJoules of energy are taken out of the soup, 00:00:16.850 --> 00:00:19.440 which is 377 times 10 to the 3 Joules. 00:00:20.567 --> 00:00:24.320 And the question is “what is the final temperature of the soup?” 00:00:25.047 --> 00:00:29.250 So, we have three heat transfer terms to consider: 00:00:29.461 --> 00:00:35.665 one is the heat transfer to change the temperature of the soup and bowl 00:00:35.898 --> 00:00:38.145 from 25 degrees Celsius to zero; 00:00:38.821 --> 00:00:40.101 and then once it’s at zero, 00:00:40.283 --> 00:00:45.294 we have the heat transfer to freeze the water from a liquid into a solid, 00:00:46.145 --> 00:00:47.614 and that’s the heat of fusion; 00:00:48.021 --> 00:00:48.894 and then after that, 00:00:49.156 --> 00:00:53.563 we have a further heat transfer to reduce the temperature from zero degrees Celsius 00:00:53.563 --> 00:00:56.334 to whatever the final temperature is that we’re going to calculate. 00:00:57.112 --> 00:01:00.610 So that’s the total energy taken out, which we’re told is this much. 00:01:01.941 --> 00:01:03.294 So some information that we, 00:01:03.294 --> 00:01:07.570 I need to research here is the latent heat of fusion for water 00:01:07.570 --> 00:01:10.494 which is 334 times 10 to the 3 Joules per kilogram. 00:01:10.952 --> 00:01:14.661 The specific heat of aluminum is 900 Joules per kilogram per Celsius degree. 00:01:14.661 --> 00:01:18.610 And the specific heat of the soup, we’re told to assume is the same as that of water, 00:01:18.989 --> 00:01:22.669 which is 4186 Joules per kilogram per Celsius degree. 00:01:23.192 --> 00:01:25.127 And the specific heat of ice, 00:01:25.340 --> 00:01:29.650 which will be needed for this final heat transfer term here, 00:01:29.847 --> 00:01:33.381 where we’re changing the temperature of the frozen soup and bowl 00:01:33.563 --> 00:01:36.298 from zero to whatever final temperature is. 00:01:37.243 --> 00:01:39.301 The specific heat of ice will be this much. 00:01:40.989 --> 00:01:46.356 Okay, and we’re told that there’s .25 kilograms of aluminum in the bowl, 00:01:46.356 --> 00:01:48.458 and .8 kilograms of soup. 00:01:51.185 --> 00:01:56.225 So, the first step is to substitute for the heat transfer 00:01:56.225 --> 00:02:00.843 to change temperature from 25 degrees Celsius to zero. 00:02:01.498 --> 00:02:07.672 Now, a small oddity is that I’ve reversed these temperature terms. 00:02:07.854 --> 00:02:09.789 Normally, you have final minus initial, 00:02:09.789 --> 00:02:12.872 and the final temperature is zero, and the initial temperature is 25. 00:02:13.170 --> 00:02:14.640 But I’ve reversed them around 00:02:15.629 --> 00:02:19.520 because we want to have the heat transferred out of the soup. 00:02:19.723 --> 00:02:23.520 This Q formula, you know, Q equals mc delta T, 00:02:23.810 --> 00:02:25.920 is the amount of heat transferred into something. 00:02:26.094 --> 00:02:28.290 And when heat is transferred out, it becomes negative. 00:02:28.465 --> 00:02:30.741 And when it becomes negative, we can have, you know, 00:02:30.741 --> 00:02:36.712 it’s negative mc T final minus T initial, if you write it the proper way. 00:02:36.916 --> 00:02:42.676 But this can be instead written as positive mc times T initial minus T final. 00:02:43.141 --> 00:02:49.047 And so, you can make this negative Q, you can write it instead like this: 00:02:49.483 --> 00:02:54.043 you have mc times the temperatures with the initial written first, 00:02:54.043 --> 00:02:55.556 and the final written second. 00:02:55.949 --> 00:02:58.109 And so that’s why this is written that way. 00:02:58.938 --> 00:03:01.890 So, we’re making this whole term positive in other words. 00:03:03.527 --> 00:03:09.272 And then we have the heat of fusion: 00:03:09.272 --> 00:03:11.927 the mass of the soup times the latent heat of fusion for water. 00:03:12.407 --> 00:03:16.116 And then add to that the mass of the aluminum 00:03:16.298 --> 00:03:17.672 times the specific heat of aluminum 00:03:17.672 --> 00:03:19.883 plus the mass of the soup times the specific heat of ice, 00:03:20.116 --> 00:03:24.043 as we go from zero degrees to some final temperature that we will calculate. 00:03:25.061 --> 00:03:30.647 So, we distribute this bracket into this one here, 00:03:31.200 --> 00:03:35.752 which conveniently turns to only two terms because... 00:03:36.225 --> 00:03:40.800 or one if depending on how you look at it because multiplied by zero makes zero. 00:03:41.127 --> 00:03:46.385 And so we have this term for the change in temperature 00:03:46.385 --> 00:03:49.156 from 25 degrees Celsius to zero written here, 00:03:49.556 --> 00:03:53.985 and that’s a minus because I subtracted it from both sides 00:03:53.985 --> 00:03:58.036 so I’m taking this to the left hand side, 00:03:58.814 --> 00:04:03.032 and then this term also is going to the left hand side 00:04:03.032 --> 00:04:05.258 or we’re subtracting it from both sides, you might say. 00:04:05.621 --> 00:04:08.370 And then we’re left with this, by itself, on one side 00:04:08.901 --> 00:04:11.294 and this bracket here multiplied by zero makes zero. 00:04:11.418 --> 00:04:14.756 And this bracket is getting multiplied by negative T final. 00:04:14.865 --> 00:04:16.167 So it’s negative T final. 00:04:16.363 --> 00:04:18.698 And then I switch the sides around as well after all that 00:04:18.923 --> 00:04:20.727 in order to have the unknown on the left. 00:04:21.556 --> 00:04:25.658 Okay, so we have negative T final times all this written here. 00:04:25.978 --> 00:04:29.520 And then equals the total energy transferred from the soup 00:04:29.890 --> 00:04:32.632 minus each of these two terms here. 00:04:35.200 --> 00:04:36.029 Okay. 00:04:37.040 --> 00:04:40.007 And then, we’ll solve for T final. 00:04:40.363 --> 00:04:43.047 We’ll divide both sides by this bracket here. 00:04:43.156 --> 00:04:51.396 So that’s divided by negative m al C al, plus m soup C ice, both sides. 00:04:58.960 --> 00:05:04.443 Now, this negative makes each of the terms in the numerator change the sign. 00:05:04.770 --> 00:05:08.785 And so that’s why we have a positive ms Lf here, 00:05:09.025 --> 00:05:13.956 and a positive bracket times T i, and then the Q t is negative. 00:05:14.843 --> 00:05:17.003 Then on the bottom we’ve written positives. 00:05:18.261 --> 00:05:20.894 Okay, then we plug in numbers and get our answer. 00:05:21.476 --> 00:05:22.647 So the final temperature is: 00:05:22.647 --> 00:05:26.872 the quarter kilograms of aluminum multiplied by specific heat of aluminum; 00:05:27.061 --> 00:05:30.763 plus the .8 kilograms of soup multiplied by specific heat of water, 00:05:30.938 --> 00:05:32.872 while it’s in the water liquid state; 00:05:33.316 --> 00:05:35.498 times 25 Celsius degrees --- 00:05:35.498 --> 00:05:37.498 change in temperature from 25 to zero; 00:05:38.000 --> 00:05:39.418 plus the heat of fusion, 00:05:39.418 --> 00:05:42.654 which is the mass of the soup multiplied by the latent heat of fusion for water; 00:05:43.250 --> 00:05:48.000 and then subtract from that the total energy taken from the soup; 00:05:48.800 --> 00:05:52.727 and then divide by the mass of aluminum times specific heat of aluminum 00:05:52.727 --> 00:05:57.032 plus the mass of the water times the specific heat of ice. 00:05:57.476 --> 00:06:01.920 And end up with a final temperature of negative 10.8 degrees Celsius. 00:06:04.174 --> 00:06:05.032 And there we go.