00:01
Okay, so heat conduction rate h here should be equal to ka times dlt over l.
00:05
K is the thermoconductivity.
00:07
A is the cross -sectional area, and delta t is a changing temperature.
00:11
L is the length of the material.
00:13
So in this question, the heat conduction rate of the copper should be equal to the sound of the heat conduction rate of the brass and the heat conduction rate of the steel, which will give us h copper is equal to h brass plus h steel.
00:26
If we expanded, we have kc times a, 10f minus tj over lc, is equal to.
00:31
Kba times tj minus tb over lb plus ks times a tj minus t .s over ls.
00:38
Okay.
00:39
So tf in this case is the temperature at the free end of copper, okay? and tj is the temperature at a junction point.
00:51
And tb is the temperature at a free end of brass, and ts is the temperature at the free end of the steel, okay? and we know lc is the length of copper rod and lb is the length of the brass, ls is the length of steel rod.
01:10
So in this question, we know that the temperature at the free end of the brass and the steel rod is both equal to 0 degrees celsius.
01:18
So now we can plug in back to the equation.
01:20
Then we have kc times tf minus tj over lc is equal to kb than tj over lb plus ks times tj over ls.
01:29
And you guys might want to where it's a now, okay? so well, as you can tell, in this equation, since both sides has a, that means a can be canceled out.
01:43
Okay? and this is how we get this equation here.
01:49
And if i do some arrangement here, i have a t -j, which is the temperature at a junction point, is equal to kc over l -c times tf over kc plus kv or lb plus ks over ls.
01:59
We know tf is equal to 100 degrees celsius, lc is equal to 13 .0 centimeter.
02:05
And kc, which is the thermoconductivity for the copper, is 385 watts per meter 10 kelvin.
02:12
Kb is equal to 109 watts per meter 10 kelvin, which is the thermoconductivity for the brass.
02:20
And lb, which is the length of the brass rod, is 18 .0 centimeter.
02:24
And we know ks, which is the thermal conductivity for the steel, is 50 .2 watts per meter 10 kelvin.
02:30
And ls is the length of the steel rock, which is about 24 .0 centimeter.
02:35
So if you're plugging all the values, we can determine the values for the temperature at a junction point.
02:43
So if a t -j is equal to 100 degrees celsius times kc over lc, which is 385 watts per meter 10 kelvin, over lc, which is 13 .0 centimeters.
03:10
So for this question, you guys can convert it to meter, okay? that's fine.
03:17
And also you guys can just leave it in a centimeter because eventually the unit centimeter will be canceled out...