00:01
So starting off with a.
00:03
To calculate heat transfer by conduction, we use three different areas for the walls, the roof, and the windows.
00:11
So we have the equation q conduction divided by t is equivalent to k -a -over -l for the walls, plus k -a -over -l for the roof, plus k -a -over -l for the windows, times t -1 -t2.
00:44
You plug in what you know, and you get 0 .023, joules divided by seconds times meters times the degree, times 410 meters squared, 20 .195 meters, plus 0 .1 times 280, divided by 0 .0 .055, plus 0 .84 times 33, divided by 6 .5 times 10 to the negative third, times 38 degrees celsius.
01:35
And this leaves you with 1 .832 times 10 to the 5th watts.
01:43
For part b, the energy binding added must both heat the air and replace the energy loss by conduction.
01:50
And so you have the equation q added is equal to q raise.
02:00
That's not an s raise...