00:01
We're asked to find the mass of steam at 100 degrees celsius that has to be added to one kilogram of ice at zero degrees celsius to yield liquid water at 30 degrees celsius.
00:13
So the mass of the ice, m sub i, is one kilogram.
00:17
The final temperature for the steam and the ice, or for the final change in temperature for the steam and the ice, is 30 degrees celsius.
00:27
So delta c is equal to 30 degrees celsius.
00:33
Okay.
00:34
So, and then also we have the latent heat here for the, for the ice, which is 3 .33 times 10 to the fifth.
00:48
And then else of v for the steam is 22 .6 times 10 to the fifth, both joules per kilogram.
00:54
Okay.
00:54
So the heat lost by the seam condensing and then cooling to 30 degrees celsius must be equal to the heat gain by the ice melting and then warming to 30 degrees celsius.
01:03
That's why delta s equals delta i is equal to 33 degrees celsius.
01:08
And then we can also say that ms of s multiplied by l sub v plus c sub w, that's the specific heat for water...