00:01
The mass of amount of ice given us 866 gram and the initial temperature of ice, we can write theta i, that is minus 10 .0 degrees celsius.
00:19
And after converting the given amount of ice into steam, the final temperature of the steam, we can write theta f equal to 1 to 6 degrees celsius.
00:33
And the specific heat capacity of ice equal to 2 .09 jol per gram degree celsius specific heat capacity of water 4 .186 jol per gram degree celsius and specific heat capacity of steam which is 2 .03 jolt per gram degree celsius and next we can can write the latent heat of fusion of ice lf equal to 3 .36 into 10 raised to 5 jule per kilogram and the latent heat of vaporization is 2 .26 into 10 raised to 5 joule per kilogram.
01:33
So next we can write the heat required to heat the ice from minus 10 degrees celsius to 0 degrees celsius which is mass.
01:44
Of ice specific heat of ice into 0 degree celsius minus theta i that is 866 into 2 .09 into 0 minus minus 10 so we will get 18 .1 kilojoule and next the heat required to melt the ice this is melt ice at 0 degree celsius which is equal to m into lf that is 0 .866 kilogram into 3 .36 into 10 raised to 5 so we will get approximately 291 into 10 raise to 3 joules which is equal to 291 kilojoules and the next is q3 that is heat energy required to raise the temperature of water from 0 degrees celsius 100 degrees that is mass into specific heat of water into 100 degrees celsius minus 0 degrees celsius.
03:02
That is 866 into 4 .186 into 100.
03:10
We will get 362 kilojoules.
03:15
Next is the heat energy required to convert water to steam at 100 degrees celsius.
03:22
That is q4m into lv...