00:01
So here we know the system is isolated.
00:02
So we can say that then the thermal energy of the steam plus the thermal energy of the ice should equal zero.
00:13
We can substitute in for either of these and say that then the mass of the steam, or mass of the steam times the specific heat capacity of water times the final temperature plus the mass of the ice times the specific heat capacity of water, multiplied by the final temperature.
00:36
This would be equal to the mass of the steam, multiplied by the latent heat of the steam, plus the mass of the steam, multiplied by the specific heat capacity of water times the temperature of the steam initially.
00:53
And in this case, we would add the mass of the ice, multiplied by the specific heat capacity of water times the initial temperature and then minus the mass of the ice times the latent heat of the ice and the final temperature would be equalling then the mass of the steam latent heat of the steam plus the mass of the steam specific heat capacity of water times the temperature of steam initially plus the mass of the ice, specific heat capacity of water, the initial temperature of the ice, minus the mass of the ice times the latent heat of the ice.
01:44
And this will all be divided by the mass of the steam specific heat capacity of water, plus the mass of the ice specific heat capacity of water.
01:55
And we can solve.
01:57
So the final temperature would then be equaling to 1 .0 kilogram...