00:01
Hi, so in this case what we are asked is how much energy do i require in order to change the temperature of 2 .4 kilograms of water from minus 11 degrees celsius to 74 degrees celsius.
00:18
And in order to do it, first of all, we need to consider all of the steps in the change of temperature.
00:25
In this case, as you can verify, we're going to have a change of state because we're starting with solid water at minus 11 degrees celsius.
00:33
So this is the initial and this is going to be the over here, initial temperature, final temperature, and over here we're going to have the amount of heat that we are adding to the system, right? heat.
00:54
So let's say that over here i'm at this point at my initial temperature, then i'm going to start providing heat to my system until i reach my fission temperature.
01:15
Over here, all of the heat that i provide is going to be employed for melting the water and making the phase transition from solid to liquid.
01:27
Then, after i have melted all of the ice that i have, then i can proceed to increase the temperature again.
01:36
So those are the three processes that we are required to consider in order to know what is the actual heat that we need to provide.
01:45
The next thing that we need from the information is the values for the heat capacities.
01:52
We are given the molar heat capacities on top of that.
01:56
So we know that this heat capacity for the solid is equal to 37 .1 joules per mole degree celsius.
02:07
We know that the heat capacity for the liquid is 75 .3 joules per mole degree celsius.
02:21
In this case, we are not going to consider the gas because, as you can see, my final temperature is below the boiling point of water, right? and the other information that we're going to have is the delta h of fusion because we're going to have a fission process for the change at zero degrees celsius.
02:43
So delta h of fusion is equal to 6 .01 kilojoules per mole.
02:55
And as you can see, all of the quantities here are molar quantities.
02:59
So we need to convert the 2 .4 kilograms of water to moles of water.
03:08
So 2 .4 kilograms, we know that in one kilogram we're going to have 1000 grams of each tool.
03:17
The molecular weight of water is 18 grams per mole of water.
03:23
So the total amount of water that we have in our system is going to be 133 .33 periodic moles.
03:45
Now we can proceed with the heat added for our calculation.
03:51
The total heat that we are going to require in order to have that change of temperature, it is going to be equal to amount of mole cp for the solid times the change of temperature for the solid.
04:09
That corresponds to this part of the curve.
04:18
To that we're going to add the amount of mole times the delta h of fusion.
04:25
Let me match it up...