In a certain time interval natural gas with energy content...

In a certain time interval, natural gas with energy content of 17000 J was piped into a house during a winter day. In the same time interval sunshine coming through the windows delivered 4000 J of energy into the house. The temperature of the house didn't change. What was $\Delta E_{thermal}$ of the house, and what was $|Q|$, the magnitude of the energy transfer between the house and the outside air? $\Delta E_{thermal} = \text{____ J}$ $|Q| = \text{____ J}$

Transcript

00:01 Because we have three ways of conduction, walls, roof, and windows, we simply have to add all three conductivities.

00:14 So q by d in this case is k a by l of the first.

00:19 There is walls plus k a by l for roof.

00:26 It's k a by l of windows times delta t.

00:33 So that is for walls, it's 0 .023 jules per meter second degree celsius and the area is 14 meters squared divided by the width of the walls is 0 .195 meters.

00:53 Now that similarly, we add the other two terms for the roof.

00:58 It's 0 .1 jule per meter.

01:00 K is 0 .1 joule per second degree celsius.

01:05 The area is to 80 meters square divided by.

01:08 The width is 0 .055 meters.

01:13 That's for windows k is 0 .84 watt per second degree celsius.

01:20 Area is 33 meters squared divided by the width is 6 .5 into 10 power minus 3 meters.

01:26 All of this needs to be multiplied by 38 degrees celsius.

01:32 The temperature difference.

01:36 It's 23 minus 15 by the way this when you substitute into our calculator turns out to be 1 .832 into 10 par 5 watts, which is about 1 .8 into 10 part 5 watts.

01:53 This is how fast the heat is dissipated.

01:58 Now as the room is heated, the temperature inside goes from 12 degrees celsius to 23 degrees celsius.

02:10 The outside temperature is minus 15 degrees celsius.

02:13 So the temperature difference goes from 12 minus minus 50, so 27 degrees celsius, to 23 minus minus 50, which is 38 degrees celsius, which means the average temperature difference corresponds to 32 .5 degrees celsius.

02:40 So we add some heat.

02:43 This heat is used to raise the temperature of the room, plus that is heat is also lost due to conduction.

03:01 So the amount of heat needed to raise the temperature is simply mc delta d, which is in this case where in this case the mass of the air is rho v, where v is the volume and r0 is the density of air and of course c is the special heat of air and delta d is the change in temperature...

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