00:01
We are given a 43 .1 gram piece of lead at initial temperature of 20 .0 degrees c.
00:13
We are told that we apply 75 .5 .4 joules of energy, and it wants to know our final temperature.
00:35
So what i'm going to do is find delta t.
00:40
Just a little easier to watch that way.
00:42
That's not normally the way i do it.
00:44
Find delta t and then i will subtract to find t f and i think i'm going to subtract what am i doing here so i'm going to have so i'm going to have my t final minus t initial equals delta t so my t final will equal i will add okay i should have left that there okay and then see i need to find and i'm going to go look for specific heat of lead.
02:02
Specific heat of lead is 0 .128 joules per kilogram of, excuse me, joules per gram degrees c.
02:21
And that should get me all ready to go here.
02:26
Okay, we'll be using q equals m times c times delta t where delta t equals q m c q is 75 .4 joules divided by 43 .1 grams divided by 0 .128 j over g degrees c...