00:01
The things we know from this problem is that we have a tungsten heating element.
00:07
It is connected to a battery with a negligible internal resistance.
00:13
And at a temperature, let's call it ti, which is equal to 80 degrees c, it draws a power, let's call that pi, of 480 watts.
00:25
We can assume that the conditions are such that we can use table 25 .2, and we're also told to use equation 25 .12 with a reference temperature t not of 20 degrees celsius.
00:45
And we want to know at a temperature, let's call that tf, of 150 degrees celsius, what is the power that is consumed by this? heating element.
01:00
So since we're told to use equation 25 .1 .2, i went ahead and wrote it down, and we're also told that we can use table 25 .2.
01:10
So this alpha over here, that coefficient corresponds to the one for tungsten, which is 0 .0045 per degree celsius.
01:28
Now, given that we have to find a power and we know that we have to use an equation with resistance, that leads me to think that i should use this equation here.
01:47
Power is equal to the potential difference squared over r, right? but r is a function of temperature, so let's write that.
02:01
So then my power becomes a function of temperature.
02:10
Now for each case, before i do that, let's rewrite this as v squared is equal to pr.
02:26
So we have this situation where you have you supply some potential difference, and then at pi, at some r .i, at some temperature i, you have p .i...