WEBVTT 00:00:00.000 --> 00:00:03.240 This is College Physics Answers with Shaun Dychko. 00:00:03.280 --> 00:00:06.340 We are going to compare the rate of heat transfer through a wall, 00:00:06.380 --> 00:00:09.080 which I have labeled with subscripts 1, 00:00:09.160 --> 00:00:11.720 to the rate of heat transfer through a window, 00:00:11.720 --> 00:00:15.100 which has all of its details labeled with a subscript 2. 00:00:15.440 --> 00:00:17.700 So the wall is 13.00 centimeters thick 00:00:17.740 --> 00:00:19.760 and since we are comparing two things... 00:00:19.760 --> 00:00:21.760 I know we are going to be dividing and so 00:00:21.760 --> 00:00:23.960 if we have units that are the same— 00:00:23.960 --> 00:00:26.040 centimeters for the thickness of the wall 00:00:26.040 --> 00:00:27.900 and centimeters for the thickness of the window— 00:00:27.900 --> 00:00:30.360 we probably don't need to change those units. 00:00:30.360 --> 00:00:32.260 We'll check to make sure when we get further along 00:00:32.300 --> 00:00:35.480 but if we are gonna divide centimeters by centimeters that will be fine; 00:00:35.480 --> 00:00:37.600 there's no need to convert this into meters. 00:00:39.120 --> 00:00:41.660 So the area of the wall is 10.0 square meters 00:00:41.660 --> 00:00:45.420 and the thermal conductivity of the wall is 2 times that of glass wool 00:00:45.420 --> 00:00:47.500 because it's probably full of fiber glass 00:00:47.500 --> 00:00:49.860 so fiber glass is another name for glass wool 00:00:50.060 --> 00:00:54.820 and so it's 0.084 joules per second per meter per Celsius degree. 00:00:55.080 --> 00:00:58.220 The thickness of the window... we assume it's just a single pane window 00:00:58.220 --> 00:01:02.240 because it has such a small thickness of 0.750 centimeters 00:01:02.300 --> 00:01:06.600 and we need to figure out what the thermal conductivity of it is 00:01:06.600 --> 00:01:08.600 and we'll have to use that of glass 00:01:08.660 --> 00:01:13.200 which is 0.84 joules per second per meter per Celsius degree 00:01:13.300 --> 00:01:15.880 and the area of the window is 2.00 meters squared. 00:01:16.680 --> 00:01:19.020 So the rate of heat conduction 00:01:19.060 --> 00:01:22.520 is the thermal conductivity multiplied by the area of something 00:01:22.520 --> 00:01:25.920 times the temperature difference between one side and the other 00:01:25.960 --> 00:01:29.700 divided by the distance between these two sides. 00:01:29.700 --> 00:01:34.980 And so for the wall that is Q over t number 1, 00:01:35.040 --> 00:01:37.900 we have the thermal conductivity of the wall times the area of the wall 00:01:37.900 --> 00:01:39.340 times this temperature difference 00:01:39.380 --> 00:01:43.280 which we don't need to label with subscripts 1 or 2 00:01:43.320 --> 00:01:45.960 because it's the same temperature difference on the inside of the wall 00:01:46.000 --> 00:01:47.420 and the outside of the wall as it is 00:01:47.420 --> 00:01:49.540 in the inside of the window and the outside of the window 00:01:49.760 --> 00:01:52.340 so they are both in contact with the same air 00:01:52.340 --> 00:01:55.060 inside the house and the same air outside the house. 00:01:56.100 --> 00:01:58.500 But the thickness of the wall is distinct 00:01:58.500 --> 00:02:01.640 and so it has a subscript 1 there for the thickness of the wall. 00:02:01.700 --> 00:02:02.480 Okay! 00:02:02.520 --> 00:02:06.640 So this is getting divided by the same expression for the window 00:02:06.640 --> 00:02:09.320 but I find it confusing to divide a fraction by a fraction 00:02:09.320 --> 00:02:13.140 and so instead I am multiplying by the reciprocal of the denominator. 00:02:13.260 --> 00:02:17.980 So this is this rate of heat conduction across the window 00:02:17.980 --> 00:02:21.160 but I flipped it over since we are dividing by it. 00:02:21.480 --> 00:02:23.180 So this is the thickness of the window 00:02:23.180 --> 00:02:25.060 divided by the thermal conductivity of the window 00:02:25.060 --> 00:02:27.820 times its area times the same temperature difference. 00:02:28.080 --> 00:02:31.440 So this bracket T a minus T b can be canceled— 00:02:31.500 --> 00:02:33.940 it's a common factor on the top and bottom— 00:02:34.020 --> 00:02:39.080 and so we are left with this expression: k 1A 1d 2 over k 2A 2d 1. 00:02:39.560 --> 00:02:42.800 So we have 0.084—thermal conductivity for the wall— 00:02:42.800 --> 00:02:44.800 times the wall area of 10.0 square meters 00:02:44.840 --> 00:02:47.820 times the window thickness of 0.750 centimeters 00:02:47.860 --> 00:02:50.820 and here we can see that we are dividing 0.750 centimeters by 00:02:50.860 --> 00:02:53.740 13.0 centimeters and since the centimeter units are the same— 00:02:53.740 --> 00:02:56.960 this is all good—we don't need to do unit conversions there. 00:02:57.380 --> 00:03:01.960 And so we are dividing by the thermal conductivity of the window—0.84— 00:03:01.960 --> 00:03:03.960 times its area of 2.00 square meters 00:03:03.960 --> 00:03:06.380 and we get 0.029. 00:03:06.660 --> 00:03:09.080 So this says the rate of heat transfer through the wall 00:03:09.100 --> 00:03:12.500 is only 2.9 percent, if you convert that into a percent, 00:03:12.540 --> 00:03:14.600 that of the window 00:03:14.640 --> 00:03:17.620 and that's quite dramatic because the wall is 00:03:17.660 --> 00:03:20.680 5 times larger area than the window 00:03:20.760 --> 00:03:26.180 and yet it has only 2.9 percent of the rate of heat loss.