00:01
Change in temperature of water that is delta t water is given by t final minus t initial so we have t final as 27 .29 minus 24 .0 .99 as t initial now when we subtract this the value which we are going to obtain is after subtracting 27 with 24 we get around 2 .3 degree celsius so this is the delta d water now moving forward this is was the first part.
00:34
Second part, we have q water to be calculated.
00:37
So that is going to be equal to mass of water, that is mw, multiplied by specific heat of water, multiplied by the temperature difference, that is t, delta t water.
00:47
So this is going to be, we have mass of water as 99 .8, multiplied by 4 .184 as the specific heat, multiplied by delta t water as 2 .3.
00:55
Then we get the value as 960 .3 .9 jule.
01:03
This is the value we have.
01:04
Obtain.
01:05
Now, the change in temperature of aluminum, the third part.
01:10
So, delta t, aluminum, is going to be equal to t final minus t initial.
01:21
So we have t final as 27 .29 and initial temperature we had is 200.
01:28
So this gives the values negative 172 .71 degree celsius.
01:34
Here the minus sign indicates the drastic drop of temperature...