00:01
Let's take a look at the first question.
00:03
Okay.
00:04
So the first question was asking us, what will be the heat that is required to melt a 0 .15 kilogram of lead at 327 .3 degrees celsius.
00:16
Well, we know that at 327 .3 degrees celsius is the melting point for a lead and a latent heat for lead.
00:27
For the lead at melting point is 24 .5 times 10tl power of 3 jr over kilogram.
00:38
And the mass followed it is 0 .0 .150 kilogram.
00:44
Okay? so the heat that required to melt such thing which is equal to m times l.
00:52
Let's say l1, okay.
00:55
And that will give us 0 .0150 kilogram times 24 .5 times 10.
01:01
To a power 3, jou, over kilogram.
01:05
And that will give us the heat that required to melt a lid is, and we see, it's 300, i mean, 3 ,675, jew, okay, that's for question one.
01:21
For question b was asking us the heat that required to evaporates the lid at temperature 1750 degrees, okay? so 1750 degrees is the evaporation point or evaporation temperature for a lid...