00:01
In this question, we have to find out what are the changes.
00:04
So first of all, let us consider the resulting temperature, that is, t0 degrees celsius.
00:12
First, let's find the values that are given to us.
00:14
Now, specific heat of water, that is s1, is equal to 1 calorie per gram degree celsius.
00:24
Then s2, specific heat of ice is 2 is equal to 0 .5 gram per degree celsius.
00:34
Now, let us consider the resulting temperature, t0, resulting temperature.
01:10
Now the heat lost by water, let's call it q1, is equal to m1 s1, del d1.
01:21
Putting the values, this is equal to 10 times, 1 times 100 minus t0, calorie...