WEBVTT 00:00:00.000 --> 00:00:03.240 This is College Physics Answers with Shaun Dychko. 00:00:03.680 --> 00:00:07.620 The Sun is a sphere with the radius of 7 times 10 to the 8 meters 00:00:07.920 --> 00:00:12.000 and it has a net radiation of 00:00:12.040 --> 00:00:14.980 3.80 times 10 to the 26 watts 00:00:15.040 --> 00:00:19.320 and we put a negative sign there to indicate that it's losing heat 00:00:19.320 --> 00:00:22.240 to the surrounding space which has a temperature of 00:00:22.240 --> 00:00:23.540 3 degrees Kelvin 00:00:23.540 --> 00:00:25.060 or I should say just 3 Kelvin... 00:00:25.100 --> 00:00:26.920 not use the word 'degrees' by the way. 00:00:27.160 --> 00:00:28.040 Okay! 00:00:28.220 --> 00:00:32.160 So the emissivity of the Sun is a perfect 1.00 it says. 00:00:32.440 --> 00:00:34.480 So we are meant to find 00:00:34.480 --> 00:00:39.220 the temperature of the Sun knowing that it has a net radiation of this amount 00:00:39.240 --> 00:00:41.600 to an environment at this temperature. 00:00:42.600 --> 00:00:46.340 So we can say that the rate of heat loss due to radiation is 00:00:46.340 --> 00:00:49.060 Stefan-Boltzmann's constant multiplied by the emissivity 00:00:49.080 --> 00:00:51.420 multiplied by the surface area of the Sun 00:00:51.420 --> 00:00:58.920 and since it's a sphere, its surface area will be 4π times the radius squared 00:00:59.340 --> 00:01:01.580 and then we'll multiply that by 00:01:01.580 --> 00:01:04.140 the temperature of the space to the power of 4 00:01:04.180 --> 00:01:06.620 minus the temperature of the Sun to the power of 4 00:01:06.660 --> 00:01:09.860 and it's this T 1 that we want to solve for. 00:01:10.480 --> 00:01:15.260 So we'll divide both sides by σ times e times A 00:01:16.920 --> 00:01:21.060 and I am gonna keep this Q over t as a single thing... 00:01:21.060 --> 00:01:25.640 it's this single number here so we'll just take this Q over t 00:01:25.640 --> 00:01:28.820 and divide the whole fraction by σeA 00:01:29.040 --> 00:01:30.900 and then switch the sides around as well. 00:01:31.580 --> 00:01:36.320 So then we wanna solve for T 1 to the power of 4, to begin with, 00:01:36.400 --> 00:01:40.280 and so add T 1 to the power of 4 to both sides 00:01:40.960 --> 00:01:44.520 and then subtract this whole fraction from both sides so it goes to the left 00:01:44.520 --> 00:01:47.040 and then switch the sides around 00:01:47.040 --> 00:01:50.580 and we end up with T 1 to the power of 4 on its own on one side 00:01:50.580 --> 00:01:53.660 and T 2 to the power of 4 minus 00:01:53.900 --> 00:01:57.300 Q over t over σeA on the other side. 00:01:58.380 --> 00:02:02.860 Now the area of a sphere is 4πr squared so we can plug that in for A 00:02:02.920 --> 00:02:05.980 and I also took the fourth root of both sides 00:02:05.980 --> 00:02:08.720 so this to the power of one quarter 00:02:08.800 --> 00:02:11.080 equals this to the power of one quarter 00:02:11.080 --> 00:02:13.660 and T 1 to the power of 4 00:02:13.660 --> 00:02:17.360 to the power of one quarter becomes T 1 to the power of 1 00:02:19.540 --> 00:02:23.100 and all of that equals the right hand side to the power of one quarter 00:02:24.260 --> 00:02:25.580 so then we plug in numbers. 00:02:25.620 --> 00:02:30.020 So that's 3.00 Kelvin to the power of 4 minus negative 3.80 times 10 to the 26 watts 00:02:30.220 --> 00:02:33.040 divided by 5.67 times 10 to the minus 8 00:02:33.080 --> 00:02:35.220 times 1 for the emissivity 00:02:35.260 --> 00:02:37.940 times 4π times 7.00 times 10 to the 8 meters— 00:02:37.940 --> 00:02:40.720 radius of the spherical Sun—squared 00:02:40.800 --> 00:02:42.580 and all that to the power of one quarter 00:02:42.620 --> 00:02:46.200 gives 5.74 times 10 to the 3 Kelvin 00:02:46.520 --> 00:02:50.200 so it's about 6000 Kelvin on the surface of the Sun. 00:02:51.460 --> 00:02:58.620 Now in part (b), we are asked how much does the Sun radiate per area 00:02:58.620 --> 00:03:00.600 or per square meter on its surface? 00:03:00.660 --> 00:03:04.160 So we take that total power that we are given at the beginning 00:03:04.160 --> 00:03:07.520 and divide it by the surface area of the Sun 00:03:07.520 --> 00:03:09.760 and that surface area is 4πr squared. 00:03:09.800 --> 00:03:12.100 So it's 3.80 times 10 to the 26 watts 00:03:12.100 --> 00:03:15.340 divided by 4π times 7.00 times 10 to the 8 meters squared 00:03:15.340 --> 00:03:19.500 and that's 6.17 times 10 to the 7 watts per square meter. 00:03:20.180 --> 00:03:24.400 And in part (c), the radius that we use in our area 00:03:24.400 --> 00:03:27.640 is going to be the distance from the Earth to the Sun now. 00:03:28.300 --> 00:03:32.380 So we imagine that the same power output is now being spread over 00:03:32.380 --> 00:03:37.180 a sphere with a radius equal to the Earth-Sun distance. 00:03:37.540 --> 00:03:41.320 So that's 3.80 times 10 to the 26 watts divided by 4π times 00:03:41.320 --> 00:03:44.460 1.514 times 10 to the 11 meters squared— 00:03:44.520 --> 00:03:46.760 this is the distance from the Earth to the Sun— 00:03:46.860 --> 00:03:50.780 and this works out to 1.32 times 10 to the 3 watts per square meter. 00:03:52.200 --> 00:03:53.960 This, by the way, has a name -- 00:03:53.960 --> 00:03:56.640 it's called the 'Solar Flux', the amount of 00:03:56.740 --> 00:04:02.140 radiative energy per second per meter squared.