In this video we're gonna look at the heat required for phase changes. The first phase
change we'll be taking ice and converting it into water. So taking solid water going
through the fusion or melting process to get the liquid water. If we reverse this go from
water to ice this is freezing. The delta H for fusion of water is 6.01 kJ/mol. Notice
that this is a molar quantity. The second phase that we'll be looking at is going from
water to steam so taking liquid water, going through the vaporization process to come up
with gaseous water. If we reverse this, go from steam to water we have condensation.
The delta H for vaporization of water is 40.79 kJ/mol again notice it's a molar quantity.
If we think about the process of taking chilled ice all the way to heated steam the first
step is going to be heating up the ice to the melting point of water 0°C. Once we're
there we're gonna have to add more energy to the process to convert the solid water
into liquid water, the heat of fusion. Then we'll heat the water up from 0°C to 100°C
and we'll have to do a vaporization of water to go from 100°C water to 200°C steam. The
last step is the heating of the steam. Let's look at a simple problem. How much energy
is required to convert 25.0 g of water at 25°C to steam at 110°C? We can break this
one into three steps. First, heating of the water 25°C to 100°C water. Vaporization
of water 100°C water to 100°C steam and then heating of the steam 100°C steam to
110°C steam. First, heating water. We're gonna use q equals mass times the specific
heat times the change in temperature. Mass in this case is gonna be 25.0 g. specific
heat of liquid water is 4.184 J/(g °C). The change in temperature is going to be 100°C
-25°C or 75°C. Plug those in for our equation and we see that q equals 25.0g multiplied
by 4.184 J/(g °C) x 75°C. Do our mathematics and we see that q is equal to 7,850 J or 7.85
kJ. Second step vaporization of water. In this case we're gonna need to know the moles
of water not the grams so therefore we're gonna take 25.0 g x 1 mol over 18.02 g. Cancel
out grams and we get 1.39 mol. We can then plug that into the formula q equals the number
of moles times the enthalpy of vaporization or n Delta H vaporization. Plug those two
numbers in and we see we get q = 1.39 mol x 40.79 kJ/mol. That gives us 56.6 kJ. The
third step heating of steam again we use q equals mass times specific heat times the
change in temperature. Mass again is 25.0 g. Specific heat this case we need to be careful
that we're using specific heat of gaseous water and not liquid water. This is 2.09 J/(g
°C). And here our temperature change is going from 100° to 210° so we see a change in
temperature of 10°C. Plug that into our formula q = 25.0 g x 2.09 J/(g °C) x 10°. Grams
and degrees Celsius cancel giving us joules. So q= 523 J or 0.523 kJ. Now we have the three
separate steps of the heating process or how much energy has gone into the system so we
just had to add 'em up together. So total energy is going to be q first step plus q
the second step plus q the third step. That gives us 7.85 kJ +56.6 kJ +0.523 kJ. Therefore
our total energy is gonna be 65.0 kJ. Now in this example we only went through three
of the processes: heating of the water, vaporization of the water, and then heating of steam. You
can have up to five steps in most cases and of course you can also have a sublimation
process where you go from solid all the way to gas. Problems aren't difficult but they
do take practice. Practice and get confident.