00:01
Hi, in this question, when coffee is cooled from 89 degrees celsius to 61 degrees celsius, the heat lost q can be given by mc delta.
00:32
Now from the question, the mass is 300 ml which is 300 grams.
00:45
So this will be 0 .3 kilograms and c is the specific heat of water which is 4 ,200 jouled per kilogram degree celsius and the change in temperature is 89 degrees celsius minus 61 degrees celsius upon substituting the values q will be equal to 0 .3 into 4 .3 4 ,200 into 89 minus 61 which will give a value of 3 5 ,2, 8, 0 joules.
01:38
Now for the given process there are 3 steps.
01:46
In the first steps, the temperature rises from minus 16 degrees celsius to 0 degrees celsius.
01:58
In the second step, the melting of ice happens at 0 degrees celsius.
02:11
And in the third step, heating of ice to 61 degrees celsius.
02:23
Now for this whole process, the heat exchange q can be given by mass of ice into specific heat of ice into the heat.
02:45
Change in temperature in the first step that is 0 minus 16 plus for the melting of ice mass into latent heat of fusion of ice plus for the third step mass into specific heat of water into the change in temperature that is it changes from 0 to 60 degrees so 60 minus 0.
03:18
Now mc, that is the mass of ice cube is common.
03:25
We'll take that outside and rearrange...