00:00:00.290 let's begin our discussion with heat 00:00:02.99000:00:03.000 transfer there's three ways in which an 00:00:05.88900:00:05.899 object could transfer heat energy that 00:00:10.43000:00:10.440 is conduction convection and radiation 00:00:14.04900:00:14.059 now let's say if we have an object that 00:00:17.90000:00:17.910 is at a relatively high temperature of 00:00:19.76000:00:19.770 100 degrees Celsius and let's say the 00:00:23.30000:00:23.310 temperature of the environment is 25 00:00:25.91000:00:25.920 degrees Celsius if you touch the object 00:00:31.55000:00:31.560 with let's say a metal rod heat will 00:00:36.08000:00:36.090 flow from the object directly to the 00:00:37.97000:00:37.980 metal whenever heat transfer occurs by 00:00:42.41000:00:42.420 contact its conduction now the object 00:00:47.33000:00:47.340 can also transfer heat by convection air 00:00:51.70000:00:51.710 can carry away heat energy from the hot 00:00:54.38000:00:54.390 object so whenever heat energy is 00:00:57.68000:00:57.690 transferred by the movement of a fluid 00:00:59.70900:00:59.719 be it a liquid or gas that's convection 00:01:03.41000:01:03.420 a good example of convection is let's 00:01:06.62000:01:06.630 say if you have a hot surface this is 00:01:13.13000:01:13.140 going to heat up the air above it so hot 00:01:16.78900:01:16.799 air will rise 00:01:18.67000:01:18.680 whenever you heat up an object it 00:01:21.13000:01:21.140 expands and density is mass over volume 00:01:26.32000:01:26.330 so when the air is heated it expands the 00:01:29.80000:01:29.810 volume decreases until I mean the volume 00:01:33.26000:01:33.270 increases I take that back as it expands 00:01:36.49900:01:36.509 and the density decreases whenever you 00:01:40.37000:01:40.380 have a light object light objects tend 00:01:42.71000:01:42.720 to float for example if you put an 00:01:44.56900:01:44.579 object that's less dense and water is 00:01:46.58000:01:46.590 going to flow an object that's more 00:01:48.67900:01:48.689 dense than water will sink so hot air 00:01:52.28000:01:52.290 because it's light it rises up to the 00:01:55.34000:01:55.350 top cold air will sink and so you have 00:02:00.38000:02:00.390 this heat transfer going on and this is 00:02:02.20900:02:02.219 an example of convection where heat is 00:02:04.78900:02:04.799 being transferred by the movement of a 00:02:06.28900:02:06.299 fluid now the last example 00:02:12.17000:02:12.180 is radiation so this hot object can 00:02:20.86900:02:20.879 radiate heat in all directions 00:02:26.47000:02:26.480 typically heat is transferred by an 00:02:28.94000:02:28.950 infrared radiation and there's no 00:02:32.78000:02:32.790 contact at all so think of the Sun the 00:02:36.19900:02:36.209 Sun emits infrared rays and visible 00:02:39.19900:02:39.209 light waves as well but that radiation 00:02:41.75000:02:41.760 comes to us without the need of anything 00:02:46.12900:02:46.139 in between because space is empty so 00:02:50.03000:02:50.040 that's the third form of energy transfer 00:02:52.00900:02:52.019 it's radiation there's no contact for 00:02:54.86000:02:54.870 this type of heat transfer so make sure 00:02:58.39900:02:58.409 you know the difference between 00:02:59.31900:02:59.329 conduction which requires contact 00:03:02.25900:03:02.269 convection which deals with the movement 00:03:06.53000:03:06.540 of a fluid like a liquid or gas or 00:03:08.69000:03:08.700 radiation which doesn't require any 00:03:11.62900:03:11.639 medium to transfer heat energy now 00:03:19.46000:03:19.470 there's an equation that can help you to 00:03:22.09900:03:22.109 calculate the rate of heat flow whenever 00:03:24.40900:03:24.419 two objects are in contact with each 00:03:26.59900:03:26.609 other so let's say if you have something 00:03:32.47900:03:32.489 that looks like this 00:03:37.86000:03:37.870 on the left side let's say the 00:03:39.86900:03:39.879 temperature is a hundred degrees Celsius 00:03:41.46000:03:41.470 and on the right side it's only twenty 00:03:44.97000:03:44.980 degrees Celsius you know heat is going 00:03:47.97000:03:47.980 to flow from hot to cold 00:03:56.83000:03:56.840 you can think of heat as being 00:03:59.32000:03:59.330 associated with the energy of the 00:04:00.85000:04:00.860 molecules the higher the temperature of 00:04:03.94000:04:03.950 an object the more kinetic energy the 00:04:06.37000:04:06.380 molecules have their vibrate and more 00:04:08.05000:04:08.060 their collide and more and so these 00:04:10.69000:04:10.700 molecular collisions will eventually 00:04:13.65000:04:13.660 transfer their energy from one place to 00:04:15.79000:04:15.800 another so heat will naturally flow from 00:04:19.72000:04:19.730 hot to cold now there's an equation that 00:04:24.03000:04:24.040 helps you to relate the amount of heat 00:04:26.83000:04:26.840 that's going to flow in any given time 00:04:28.75000:04:28.760 period so you need to know that the rate 00:04:34.36000:04:34.370 of heat flow is proportional to the area 00:04:36.87900:04:36.889 the cross-sectional area of the 00:04:39.65900:04:39.669 conductor it also is dependent on the 00:04:45.04000:04:45.050 temperature difference if these two 00:04:48.79000:04:48.800 temperatures are the same there's going 00:04:50.74000:04:50.750 to be no heat flow the greater the 00:04:53.74000:04:53.750 temperature difference the greater the 00:04:55.65900:04:55.669 amount of heat flow that will occur so 00:04:59.26000:04:59.270 the rate of heat flow depends on the 00:05:00.52000:05:00.530 change of temperature and also the 00:05:03.07000:05:03.080 length between the hot and the cold 00:05:07.15000:05:07.160 section so if you increase that distance 00:05:11.08000:05:11.090 that should decrease the rate of the 00:05:13.48000:05:13.490 heat flow here's the equation that you 00:05:16.99000:05:17.000 need to know Q which represents the 00:05:19.90000:05:19.910 amount of heat energy the change in Q 00:05:22.69000:05:22.700 divided by T so that's the rate of heat 00:05:26.52900:05:26.539 energy being transferred that's equal to 00:05:29.32000:05:29.330 K which is the thermal conductivity of 00:05:31.71900:05:31.729 the substance or the material that you 00:05:35.17000:05:35.180 use in times the cross sectional area 00:05:38.15900:05:38.169 times the difference in temperature 00:05:42.65900:05:42.669 divided by L so now let's think about 00:05:47.44000:05:47.450 what this means by the way what equation 00:05:52.33000:05:52.340 what variable relates energy divided by 00:05:56.62000:05:56.630 time you need to know that power is 00:05:59.92000:05:59.930 equal to energy over time energy has 00:06:03.40000:06:03.410 units joules time is in seconds so you 00:06:07.81000:06:07.820 need to know that one watt which is the 00:06:09.77000:06:09.780 of power that's equal to one Joule per 00:06:11.90000:06:11.910 second if we increase the value of K 00:06:20.05000:06:20.060 which is the thermal conductivity the 00:06:23.54000:06:23.550 rate of heat flow will increase so this 00:06:27.08000:06:27.090 which is equal to P that's going to 00:06:29.93000:06:29.940 increase now the rate of heat flow is 00:06:33.90900:06:33.919 also proportional to the cross sectional 00:06:36.26000:06:36.270 area if you increase a the power or the 00:06:40.25000:06:40.260 rate of heat flow will increase as well 00:06:43.72000:06:43.730 basically anything in the numerator is 00:06:46.01000:06:46.020 going to be proportional to the quantity 00:06:48.74000:06:48.750 on the left side in this case the power 00:06:49.94000:06:49.950 so if we increase the change in 00:06:51.83000:06:51.840 temperature the rate of heat flow will 00:06:54.38000:06:54.390 increase as well but notice that it's 00:06:57.92000:06:57.930 inversely related to L if you increase L 00:07:01.60000:07:01.610 the energy transferred per second will 00:07:04.21900:07:04.229 decrease let's try this problem so we 00:07:11.18000:07:11.190 have a glass window let's see if I can 00:07:14.77000:07:14.780 draw a nice glass window 00:07:33.26000:07:33.270 and the glass window is 1.4 centimeters 00:07:36.47000:07:36.480 thick it has a length of 2 meters and a 00:07:47.21000:07:47.220 width of 3 meters and let's say the 00:07:51.34000:07:51.350 inside temperature is 25 degrees Celsius 00:07:55.85000:07:55.860 and the outside temperature is 22 so we 00:08:00.65000:08:00.660 know heat is going to flow from the 00:08:02.81000:08:02.820 inside to the outside from hot to cold 00:08:07.33000:08:07.340 now what we want to know is what is the 00:08:09.62000:08:09.630 rate of heat flow through a glass window 00:08:14.17000:08:14.180 now we're given the thermal conductivity 00:08:16.25000:08:16.260 of a typical glass window so how can we 00:08:20.78000:08:20.790 find a rate of heat flow now let's 00:08:28.55000:08:28.560 calculate Delta Q over T that represents 00:08:31.79000:08:31.800 the rate of heat flow which is basically 00:08:34.39000:08:34.400 the power in units of watts so using the 00:08:39.68000:08:39.690 equation we know it's K times a times 00:08:42.98000:08:42.990 the change in temperature divided by L 00:08:47.26000:08:47.270 so let's get rid of this so K which is 00:08:54.65000:08:54.660 the thermal conductivity 00:08:55.85000:08:55.860 that's 0.8 for joules per second per 00:09:01.79000:09:01.800 meter per Celsius now the area is 2 by 3 00:09:09.47000:09:09.480 so it's 2 meters by 3 meters which is 6 00:09:13.03000:09:13.040 square meters and the change in 00:09:16.40000:09:16.410 temperature if you're not sure whether 00:09:18.95000:09:18.960 the temperature should be in Celsius or 00:09:20.75000:09:20.760 Kelvin you can go with Kelvin but 00:09:23.39000:09:23.400 anytime you have a change in temperature 00:09:24.82000:09:24.830 Celsius can work so the change in 00:09:27.62000:09:27.630 temperature is 3 degrees Celsius 25 00:09:30.02000:09:30.030 minus 22 divided by the length L is 00:09:37.97000:09:37.980 basically the thickness of the glass 00:09:40.03000:09:40.040 however we do need to convert it to 00:09:42.65000:09:42.660 meters one point 100 centimeters is 00:09:46.43000:09:46.440 basically 00:09:46.91000:09:46.920 one meter so you need to divide 1.4 00:09:49.06900:09:49.079 centimeters by 100 which is point zero 00:09:51.65000:09:51.660 one four meters so notice the units 00:09:56.44000:09:56.450 Celsius will cancel and square meters 00:10:00.17000:10:00.180 will cancel with these two so we're left 00:10:03.74000:10:03.750 with the unit joules per second which is 00:10:06.62000:10:06.630 basically Watts so it's going to be 00:10:12.74000:10:12.750 point eighty four times six times three 00:10:15.88900:10:15.899 divided by point zero one four so the 00:10:19.25000:10:19.260 answer is 1080 watts so basically every 00:10:27.50000:10:27.510 second a thousand and eighty joules of 00:10:29.99000:10:30.000 heat energy is being transferred to this 00:10:33.86000:10:33.870 glass window let's try this problem if a 00:10:43.46000:10:43.470 thousand eighty joules of heat energy is 00:10:45.53000:10:45.540 transferred through the window every 00:10:46.93900:10:46.949 second how much energy will be 00:10:48.88900:10:48.899 transferred in a month so we know that 00:10:51.76900:10:51.779 power is equal to energy divided by time 00:10:54.35000:10:54.360 so the energy is the power multiplied by 00:10:57.07900:10:57.089 the time the time is one month but we 00:11:00.76900:11:00.779 need to convert answer seconds on 00:11:06.23000:11:06.240 average there's 30 days per month and 00:11:10.84000:11:10.850 it's 24 hours in a day and there's about 00:11:17.13900:11:17.149 60 minutes in an hour and 60 seconds in 00:11:20.12000:11:20.130 a minute 00:11:20.48000:11:20.490 so there's 3,600 seconds in a single 00:11:23.68900:11:23.699 hour so if we multiply 30 times 24 times 00:11:29.41000:11:29.420 3600 you should get a time value of 2 00:11:34.10000:11:34.110 million 592 seconds in a single month so 00:11:39.88900:11:39.899 now let's calculate the energy so we 00:11:43.28000:11:43.290 have a thousand 80 joules of energy 00:11:47.72000:11:47.730 being transferred every second so if we 00:11:50.21000:11:50.220 multiply by this many seconds we should 00:11:54.31900:11:54.329 get the total energy transferred in a 00:11:57.62000:11:57.630 single month 00:11:59.99000:12:00.000 so the final answer is about two point 00:12:03.99000:12:04.000 seven nine nine times 10 to the ninth 00:12:10.92000:12:10.930 power so that's how many joules is being 00:12:15.63000:12:15.640 transferred through this window in a 00:12:17.04000:12:17.050 single month now let's move on to 00:12:19.92000:12:19.930 another topic sometimes you might have 00:12:27.78000:12:27.790 to deal with our values the r-value 00:12:32.24000:12:32.250 basically describes the thermal 00:12:34.56000:12:34.570 resistance of an insulator it's equal to 00:12:39.90000:12:39.910 L divided by K where L is the thickness 00:12:47.73000:12:47.740 of the material which is the distance 00:12:52.26000:12:52.270 between the hot and cold section of the 00:12:55.05000:12:55.060 material so notice that K and R are 00:12:59.79000:12:59.800 inversely related 00:13:00.86000:13:00.870 whenever K goes up R goes down and R and 00:13:07.89000:13:07.900 L are directly related if L goes up R 00:13:11.88000:13:11.890 goes up so whenever you increase the 00:13:14.85000:13:14.860 thickness of material you increase its 00:13:17.97000:13:17.980 insulation its thermal resistance if you 00:13:22.77000:13:22.780 increase the conductivity while the 00:13:25.32000:13:25.330 insulation decreases so can are are 00:13:28.53000:13:28.540 inversely related so for example let's 00:13:36.99000:13:37.000 consider a metal such as silver metals 00:13:44.07000:13:44.080 are excellent conductors of heat and so 00:13:48.66000:13:48.670 for silver the thermal conductivity is 00:13:51.15000:13:51.160 very high it's 420 joules per second per 00:13:55.29000:13:55.300 meter per Celsius in the case of glass 00:14:00.71000:14:00.720 glass has a thermal conductivity of 0.8 00:14:05.40000:14:05.410 for 00:14:11.34000:14:11.350 fiberglass has a thermal conductivity 00:14:15.28000:14:15.290 value of point zero four eight now we 00:14:21.82000:14:21.830 can see why silver which is a metal is 00:14:25.14000:14:25.150 an excellent conductor of heat metals 00:14:29.62000:14:29.630 conduct heat very well due to the free 00:14:32.47000:14:32.480 flowing electrons that are found in 00:14:33.91000:14:33.920 metals so it has the highest K value 00:14:36.39000:14:36.400 therefore silver should have a very low 00:14:39.36000:14:39.370 r-value now glass is an insulator that 00:14:45.28000:14:45.290 doesn't really conduct heat very well 00:14:46.89000:14:46.900 but fiberglass is a better insulator 00:14:49.72000:14:49.730 because it has a lower kVAL which means 00:14:52.36000:14:52.370 fiberglass has a relatively high r-value 00:14:56.82000:14:56.830 so just keep this in mind k describes 00:14:59.53000:14:59.540 the thermal conductivity of a material 00:15:01.74000:15:01.750 so if it has a high K value it's a heat 00:15:05.17000:15:05.180 conductor if it has a low K value it's 00:15:07.09000:15:07.100 an insulator the r-value describes 00:15:10.45000:15:10.460 insulation a substance with a low 00:15:13.96000:15:13.970 r-value is a good heat conductor and the 00:15:17.35000:15:17.360 substance with a high r-value is a good 00:15:20.11000:15:20.120 insulator now let's work on this problem 00:15:25.63000:15:25.640 the r-value for a certain building 00:15:27.91000:15:27.920 material that is two point four 00:15:29.50000:15:29.510 millimeters thick is one point five what 00:15:32.02000:15:32.030 is the R value if the thickness is 00:15:33.64000:15:33.650 increased to four point eight 00:15:35.05000:15:35.060 millimeters we know that R is directly 00:15:40.60000:15:40.610 related to Al but inversely related to K 00:15:43.50000:15:43.510 so if we increase out R is going to 00:15:46.21000:15:46.220 increase now let's say if we double the 00:15:49.72000:15:49.730 value of L because it increased from two 00:15:52.48000:15:52.490 point four to four point eight and 00:15:54.91000:15:54.920 increase by a factor of two what should 00:15:58.87000:15:58.880 happen to the r-value because R is 00:16:02.86000:16:02.870 proportional to L to the first power the 00:16:05.28000:16:05.290 r-value should double as well so it 00:16:08.44000:16:08.450 should equal three point oh now we can 00:16:12.76000:16:12.770 come up with an equation that relates R 00:16:14.53000:16:14.540 now let's write the ratio between r2 and 00:16:19.15000:16:19.160 r1 00:16:20.60000:16:20.610 artoo is going to be l 2 divided by K 00:16:23.28000:16:23.290 and R 1 is L 1 divided by K for the same 00:16:27.48000:16:27.490 substance K is the same so we don't need 00:16:29.70000:16:29.710 to write a subscript for it which means 00:16:31.95000:16:31.960 we can cancel K and then we're going to 00:16:34.86000:16:34.870 get this equation r2 divided by r1 is 00:16:37.29000:16:37.300 equal to L 2 divided by L 1 so R 1 is 00:16:46.52000:16:46.530 1.5 we're looking for r2 l1 is two point 00:16:51.06000:16:51.070 four millimeters l2 is four point eight 00:16:54.26000:16:54.270 so if we cross multiply this is going to 00:16:57.48000:16:57.490 be two point four times r2 and that's 00:17:00.63000:17:00.640 equal to one point five times four point 00:17:05.61000:17:05.620 eight which is seven point two and to 00:17:09.54000:17:09.550 find our two is to divide both sides by 00:17:11.31000:17:11.320 two point four seven point two divided 00:17:13.26000:17:13.270 by two point four is three so now we can 00:17:18.60000:17:18.610 answer the second part of the problem 00:17:24.41000:17:24.420 let's use the same equation so if we 00:17:27.60000:17:27.610 decrease the thickness the R value 00:17:29.85000:17:29.860 should be less than one point five so R 00:17:33.57000:17:33.580 1 is still 1 point five we're looking 00:17:35.13000:17:35.140 for are two again l1 is two point four 00:17:38.31000:17:38.320 but l2 is one point seven so let's cross 00:17:41.49000:17:41.500 multiply one point five times one point 00:17:44.40000:17:44.410 seven is about two point 55 and that's 00:17:49.32000:17:49.330 equal to two point four r2 so now two 00:17:53.34000:17:53.350 point 55 divided by two point four will 00:17:55.86000:17:55.870 give us an r2 value of one point zero 00:17:58.50000:17:58.510 six so as you can see it's the less than 00:18:03.63000:18:03.640 one point five now the next thing that 00:18:08.40000:18:08.410 we need to quantify is the amount of 00:18:10.89000:18:10.900 heat energy that flows by means of 00:18:14.40000:18:14.410 radiation so let's say if we have an 00:18:16.89000:18:16.900 object 00:18:21.40900:18:21.419 every object emits thermal radiation as 00:18:25.18000:18:25.190 you increase the temperature of the 00:18:27.25900:18:27.269 object the more thermal radiation it's 00:18:30.82900:18:30.839 going to emit and the equation that you 00:18:33.10900:18:33.119 need it's called the Stefan Boltzmann 00:18:35.50900:18:35.519 equation the rate of heat flow is equal 00:18:41.41900:18:41.429 to the emissive the emissivity there we 00:18:45.46900:18:45.479 go 00:18:45.79900:18:45.809 almost pronounce that wrong times the 00:18:50.08900:18:50.099 Stefan Boltzmann constant times the area 00:18:53.71900:18:53.729 of the object times the temperature 00:18:57.94900:18:57.959 raised to the fourth power so this is 00:19:00.82900:19:00.839 how much energy that is being emitted by 00:19:03.52900:19:03.539 this particular object the emissivity 00:19:07.36900:19:07.379 factor varies between 0 to 1 an object 00:19:14.53900:19:14.549 with an emissivity of 1 is an object 00:19:18.28900:19:18.299 that is a good absorber of heat and also 00:19:21.34900:19:21.359 it can emit heat very efficiently if the 00:19:25.36900:19:25.379 emissivity factor is close to zero that 00:19:28.03900:19:28.049 means that it really is not good in 00:19:30.31900:19:30.329 emitting heat energy now this particular 00:19:35.11900:19:35.129 value is 5.67 00:19:39.90900:19:39.919 times 10 to the minus 8 watts per square 00:19:45.70900:19:45.719 meter per Kelvin to the fourth power so 00:19:50.98900:19:50.999 that's the Stefan Boltzmann constant and 00:19:52.87900:19:52.889 a is the area in square meters and 00:19:55.95900:19:55.969 temperature notice that we don't have a 00:19:58.31000:19:58.320 change in temperature temperature has to 00:20:00.94900:20:00.959 be in Kelvin not Celsius for this to 00:20:02.62900:20:02.639 work in this problem we have a sphere 00:20:11.44000:20:11.450 with a radius of 25 centimeters which is 00:20:17.79900:20:17.809 0.25 meters 00:20:22.98000:20:22.990 and this fear is at a temperature of 27 00:20:27.21000:20:27.220 degrees Celsius the temperature of the 00:20:30.48000:20:30.490 surroundings is positive two degrees 00:20:33.54000:20:33.550 Celsius so heat is going to flow out of 00:20:36.96000:20:36.970 the sphere into the surroundings 00:20:42.15000:20:42.160 however this heat flow is the net heat 00:20:45.63000:20:45.640 flow in this problem we need to 00:20:47.46000:20:47.470 understand that he not only leaves the 00:20:50.43000:20:50.440 sphere but also enters this view however 00:20:55.53000:20:55.540 the amount of heat that's emitted by the 00:20:58.08000:20:58.090 sphere is greater than the amount of 00:21:00.41900:21:00.429 heat that is absorbed by the sphere from 00:21:03.90000:21:03.910 the surroundings so that there's a net 00:21:06.00000:21:06.010 heat flow of heat leaving the sphere so 00:21:10.08000:21:10.090 generally speaking heat will flow from 00:21:11.61000:21:11.620 hot to cold even though goes both ways 00:21:14.43000:21:14.440 the net heat flow is from hot to cold 00:21:18.04900:21:18.059 now let's work on part a so let's move 00:21:21.87000:21:21.880 this somewhere else so what is the rate 00:21:26.49000:21:26.500 of heat energy leaving the sphere so 00:21:28.68000:21:28.690 let's use the equation Delta Q divided 00:21:31.35000:21:31.360 by delta T is equal to e theta times a 00:21:36.29900:21:36.309 times the change or just times T raised 00:21:39.60000:21:39.610 to the fourth power so the emissivity 00:21:43.29000:21:43.300 factor in this problem is 0.42 00:21:47.93000:21:47.940 the constant is always going to be the 00:21:50.58000:21:50.590 same and that is five point six seven 00:21:57.14000:21:57.150 times ten to the minus eight now what 00:22:03.27000:22:03.280 about the area of the sphere the area is 00:22:06.18000:22:06.190 equal to four pi times R squared so 00:22:11.24000:22:11.250 point twenty five squared 00:22:13.49000:22:13.500 make sure R is in meters times four pi 00:22:16.26000:22:16.270 is about 0.785 now we need to multiply 00:22:24.51000:22:24.520 it by the temperature of the sphere 00:22:26.31000:22:26.320 which is 27 degrees Celsius but we need 00:22:28.79900:22:28.809 to convert it to Kelvin to find the 00:22:31.04900:22:31.059 temperature in Kelvin add the Celsius 00:22:33.15000:22:33.160 temperature plus 273 00:22:36.24000:22:36.250 so this is going to be 300 Kelvin raised 00:22:38.97000:22:38.980 to the fourth power 300 raised to the 00:22:42.60000:22:42.610 fourth power is a big number 00:22:43.89000:22:43.900 it's about 8.1 billion if we multiply 00:22:50.82000:22:50.830 that by point seven eight five and point 00:22:56.31000:22:56.320 four two and the other number you should 00:23:02.31000:23:02.320 get a value of 150 one point four watts 00:23:08.34000:23:08.350 so every second the sphere is emitting 00:23:15.59000:23:15.600 150 one point four joules of energy so 00:23:20.19000:23:20.200 that's how much energy its emitting by 00:23:22.44000:23:22.450 radiation 00:23:23.22000:23:23.230 now how much energy is going into it by 00:23:26.34000:23:26.350 radiation we need to use the same 00:23:29.43000:23:29.440 equation by the way let's put a negative 00:23:31.91900:23:31.929 sign since this is how much energy to 00:23:36.02900:23:36.039 lose in every second now Part B to find 00:23:40.47000:23:40.480 the energy that's entering into the 00:23:42.57000:23:42.580 sphere we need to use the temperature of 00:23:44.66900:23:44.679 the surroundings so everything is going 00:23:46.59000:23:46.600 to be the same the 0.35 the constant and 00:23:50.29900:23:50.309 the area is going to stay the same 00:23:59.83000:23:59.840 the only difference is the Kelvin 00:24:01.38900:24:01.399 temperature so we need to add 273 to 2 00:24:05.04900:24:05.059 degrees Celsius to get a Kelvin 00:24:06.45900:24:06.469 temperature of 275 raised to the 4th 00:24:09.84900:24:09.859 power so if we multiply everything 00:24:20.27000:24:20.280 I do need to fix something I don't know 00:24:22.01000:24:22.020 why I have point three five that should 00:24:24.68000:24:24.690 be a point for tune 00:24:36.09000:24:36.100 so this is equal to a hundred and 6.9 00:24:40.92000:24:40.930 watts so that's the answer to Part B so 00:24:49.59000:24:49.600 in Part A the amount of energy that the 00:24:52.38000:24:52.390 sphere is losing is 150 1.4 watts Part B 00:24:59.88000:24:59.890 the amount of energy that the sphere 00:25:03.30000:25:03.310 gains per second is positive 106 point 9 00:25:08.97000:25:08.980 watts which means that there's a net 00:25:13.14000:25:13.150 loss if we combine these two 00:25:24.38000:25:24.390 of 44.5 watts so what this means is that 00:25:31.28000:25:31.290 every second the sphere loses 44.5 00:25:35.51000:25:35.520 joules of energy every second and we 00:25:39.17000:25:39.180 could see why is losing energy the 00:25:41.09000:25:41.100 temperature of the sphere is greater 00:25:43.49000:25:43.500 than the temperature of the surroundings 00:25:45.47000:25:45.480 so that means that the sphere emits more 00:25:48.71000:25:48.720 radiant energy than the amount of energy 00:25:52.04000:25:52.050 it absorbs from the environment but when 00:25:54.92000:25:54.930 the temperature of the environment is 00:25:56.27000:25:56.280 greater than the temperature of the 00:25:57.44000:25:57.450 sphere the environment will emit more 00:25:59.93000:25:59.940 radiation into the sphere than the 00:26:02.03000:26:02.040 amount that the sphere will already into 00:26:04.10000:26:04.110 the environment so whichever object has 00:26:08.18000:26:08.190 the higher temperature it's going to 00:26:09.80000:26:09.810 emit a greater amount of heat then the 00:26:13.55000:26:13.560 amount of heat that is absorbed by the 00:26:15.20000:26:15.210 surroundings so with these problems you 00:26:18.68000:26:18.690 have to think of the like how much 00:26:20.81000:26:20.820 energy is going in how much is going out 00:26:22.73000:26:22.740 and what's the net flow of the energy so 00:26:25.82000:26:25.830 this is the answer for Part C now what 00:26:28.73000:26:28.740 about Part D so we know the change in 00:26:31.88000:26:31.890 power is forty four point five watts how 00:26:38.45000:26:38.460 long will it take for the sphere to lose 00:26:41.44000:26:41.450 1 million joules of energy so remember 00:26:47.93000:26:47.940 one why is one Joule per second so the 00:26:51.32000:26:51.330 sphere it loses forty five point forty 00:26:54.68000:26:54.690 four point five joules per second you 00:26:59.39000:26:59.400 know that power is equal to energy 00:27:01.13000:27:01.140 divided by time and energy is power 00:27:04.22000:27:04.230 multiplied by time in this problem we're 00:27:06.77000:27:06.780 looking for the time the time is the 00:27:09.74000:27:09.750 energy divided by the power so let's 00:27:13.28000:27:13.290 start with the energy which is 1 times 00:27:14.81000:27:14.820 10 to the six joules and we're going to 00:27:17.63000:27:17.640 divide it by the power which is forty 00:27:21.95000:27:21.960 four point five joules per second notice 00:27:26.72000:27:26.730 that the unit joules cancel so we simply 00:27:30.47000:27:30.480 have to divide these two numbers one 00:27:32.27000:27:32.280 times ten to the six divided by forty 00:27:34.82000:27:34.830 four point five is about 00:27:37.92000:27:37.930 twenty two thousand four hundred and 00:27:40.32000:27:40.330 seventy two seconds now how many hours 00:27:45.99000:27:46.000 is that because that's a long time in 00:27:47.82000:27:47.830 seconds whenever you have a time value 00:27:52.92000:27:52.930 that's very high in seconds converted to 00:27:55.05000:27:55.060 a more practical unit so let's convert 00:27:57.78000:27:57.790 it to hours we know that there's 3,600 00:28:02.76000:28:02.770 seconds in a single hour so this is 00:28:11.49000:28:11.500 about 6.2 hours so that's how long it's 00:28:16.26000:28:16.270 going to take for the sphere to lose 1 00:28:19.71000:28:19.720 million joules of energy assuming of 00:28:21.30000:28:21.310 course if the rate of heat flow is 00:28:23.49000:28:23.500 constant that is if the temperature of 00:28:25.74000:28:25.750 the sphere and the temperature of the 00:28:27.66000:28:27.670 surroundings if it stays at that level 00:28:31.01000:28:31.020 now we know in reality the temperature 00:28:33.48000:28:33.490 of the sphere is going to decrease and 00:28:35.13000:28:35.140 as that happens the rate of the rate of 00:28:38.73000:28:38.740 heat flow will decrease as well because 00:28:40.41000:28:40.420 it's depend on temperature so as the 00:28:43.44000:28:43.450 temperature decreases the rate of heat 00:28:46.86000:28:46.870 flow will decrease as well so it's not 00:28:48.48000:28:48.490 really constant but we're going to 00:28:51.72000:28:51.730 assume as if it's constant if it is it's 00:28:54.96000:28:54.970 about 6.2 hours but in reality since the 00:28:59.31000:28:59.320 temperature is decreasing over time the 00:29:01.74000:29:01.750 rate of heat energy that's leaving the 00:29:03.48000:29:03.490 sphere will decrease and so it's going 00:29:05.88000:29:05.890 to take longer for the spheres to lose a 00:29:09.39000:29:09.400 million joules of energy so it's really 00:29:11.94000:29:11.950 longer or greater than 6.2 hours but 00:29:16.23000:29:16.240 we're going to keep this problem simple 00:29:17.43000:29:17.440 we're going to assume that the rate of 00:29:19.47000:29:19.480 heat flow stays constant
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