00:03
Here in this problem given a 638 gram block of lead at 300 kelvin absorbed 24 joules.
00:12
Specific heat capacity of lead given here.
00:15
We have to find out the final temperature.
00:19
Now recall the formula q is equal to m, c, delta, t.
00:24
Q is the heat.
00:25
M is the mass of the substance, specific heat capacity of that substance delta t, difference in temperature.
00:32
Here we will calculate delta t, that will be q divided by mc.
00:38
Here, q, that is the heat absorbs 0 .44 jule and mass of that substance 638 gram.
00:50
Time specific heat capacity, this one, 0 .74 jule per gram per degree celsius.
00:58
We can sell gram and gram, jule and jule...