00:01
In this question, we are given that the energy stored in a 52 micro -farrant capacitor is used to melt 6 milligram sample of lead.
00:17
So now we write down the information, 6 times 10 to the power negative 3 or negative 6 kg.
00:32
Okay, and then so we are also given that the initial temperature of lead is 20 degrees c, the melting temperature is 327 .3 degrees celsius.
00:55
The specific heat capacity of lead is 128 jou per kg degrees celsius.
01:06
And the latent heat of fusion is 24 .5 kilojoules per kg.
01:19
So to do this question, first we need to calculate the energy needed to melt lead.
01:32
So we'll be using certain equations such as q equals to mc, delta t, and q equals to mlf.
01:41
So one is for raising the temperature, one is for melting.
01:46
So the total energy needed to melt, lead is equal to mc delta t plus mlf.
02:03
So we can factor out the m, okay, times c delta t plus lf...