00:02
In this question, we are given a canoe engine.
00:11
It operates at th, it goes to 100 degrees celsius, and tc is 20 degrees celsius.
00:26
And then in each second, we are given that the qc.
00:34
Okay, energy exiles to the low reservoir.
00:37
At the rates of 15 .4 watts, so meaning that in each second qc is 15 .4 juice.
00:47
Okay, so there are two parts in this question.
00:49
In part a, we need to determine the useful power output.
00:55
Okay, so i'm going to solve the problem by considering the energy flow in one, second okay later we can find the useful power outputs of the heat engine okay so the efficiency is 1 minus tc over th okay so you can calculate this to be 1 minus 293 devoured by 373 and you get 0 .214 okay and actually you don't need the efficiency but okay, we can treat away use one relationship to find qh.
01:56
Okay, so we have tc over th is equal to qc, devised by qh.
02:06
So it means that qh is equal to th divided by tc times qc.
02:19
So this is equal to 373, divide by tc, 293 and 15 .4.
02:26
So this is equal to 19 .6 juice.
02:30
So w engine is qh minus qc...