00:01
First of all for water consider water first of all.
00:06
So, we know the relation that density is equals to mass divided by volume.
00:11
So, for water density of water is equals to mass of water divided by volume of water.
00:16
So, now, if i want to calculate the mass.
00:20
So, this will be density multiplied by volume given density for water is 1 .0 gram per milliliter the volume we have 352 milliliter now solving this we have 352 gram.
00:38
Now, we will apply here the specific heat capacity for water.
00:46
So, now, we know the relation that q is equals to c multiplied by m multiplied by delta t or we can write that q is equals to c multiplied by m multiplied by t2 minus t1.
01:01
So, here specific heat capacity that is the amount of heat energy we require so that there could be a temperature change by 1 degree celsius for given 1 gram of substance and here q is the amount of heat c is the specific heat capacity m is the mass and delta t is the temperature change which can be written as final temperature minus initial temperature.
01:22
Now, this is q is equals to now for water the specific heat capacity is 4 .184 joule per gram per degree celsius multiplied by mass we have 352 gram multiplied by now.
01:41
So, we have final temperature 5 degree celsius initial temperature is 25 degree celsius solving this q is equals to minus this is 29455 .36 joule...