Hello, this is a tutorial on heating curves
and calculating energy changes in a substance
undergoing a gain or loss of heat energy.
So you may see a graph like this in your textbook, called a heating curve, a temperature versus energy graph of a substance being heated. When heat is added the temperature of a substance changes in sections that are broken up by phase changes.
The energy change is referred to as delta H, meaning enthalpy change. Enthalpy itself is not measurable, but the enthalpy change is measurable because it is equivalent to a change in heat energy when pressure is constant, and we can assume constant pressure under standard laboratory conditions.
So the question we will be addressing in this tutorial is:
Let’s look at each section of the graph as energy is added.
This is the solid phase absorbing heat.
then melting occurs, then the liquid phase,
boiling, and finally the gas phase.
During melting, there is no temperature change and so
the graph remains flat, and the same occurs during boiling.
We will break it down step by step and do two example problems.
There are only two equations we will have to deal with. The first gives the energy change for any temperature change within any given phase: solid, liquid, or gas.
Delta H is equal to the mass of the substance
times the specific heat of each phase (solid, liquid, or gas)
multiplied by the change in temperature.
The specific heat, with the symbol capital C,
is the amount of energy required to raise the temperature of 1 gram of a substance by 1 degree Celsius.
For example, 1 gram of water requires 4.184 joules
to increase its temperature by 1 degree Celsius.
This may sound familiar to you:
the specific heat of water defines 1 calorie;
in other words it takes one calorie of energy to raise 1 gram of water by 1 degree Celsius.
But here we will be using the SI unit joule for energy.
The second equation we will use gives the energy change occurring during a phase change;
melting, freezing, boiling, or condensation.
The equation is heat of phase change times the number of moles.
The heat of phase change refers to the energy required for one mole of a substance to change from one phase to another.
For example, it takes 6010 joules to melt one mole of ice. The energy of melting is referred to as heat of fusion.
Let’s keep the two equations in a safe place for reference as we go through the energy changes when heating a substance,
keeping in mind that the first equation will be used when a change in temperature occurs, and the second equation is used when there is a phase change.
The first part of the graph gives
the temperature change of a solid absorbing heat,
and so of the two equations we use the equation incorporating temperature change.
This is the first calculation.
When adding more heat energy at the melting point something interesting happens.
No temperature change occurs.
If we look at what is happening during melting
we see that energy is used to break attractive forces between particles.
This results in a positional change in the particles,
and therefore it is a change in potential energy,
and without a change in kinetic energy, there will be no change in temperature.
The absorbed heat is converted to potential energy.
It is no longer kinetic energy, so no delta t.
So here we require the second equation,
using the substance’s heat of fusion, for our second delta H.
The third part of the curve is the liquid changing temperature.
And so we use the specific heat of the liquid in the third calculation.
Next is boiling.
We can see again that heat energy is used to break attractive forces.
And here they are completely broken as the substance becomes a gas,
again resulting in a positional change in the particles.
And so this is a potential energy change.
There is no change in kinetic energy,
and therefore no change in temperature.
Here we use the second equation with the substance’s heat of vaporization,
which is the energy required for a liquid
to break apart into the gas phase.
This is our fourth delta H.
Finally, the last energy change
produces a temperature change in the gas phase.
And so the specific heat of the gas is used in this fifth calculation.
Let’s go back to our original question:
Well the total enthalpy change of any substance being heated
will be the sum of the delta H of any temperature change occurring
plus the delta H of any phase change occurring.
Let’s do the first of two example problems:
What is the total enthalpy change for 10.0 grams of water
in which the temperature changes from -20 °C to +50 °C?
Since ice melts at zero degrees,
we are starting with solid water at -20,
which will heat to melting, it will melt,
and then heat to +50.
The values that we need, therefore, are the specific heat of ice,
the heat of fusion giving the energy needed to melt one mole of solid water,
and finally the specific heat of liquid water.
These values would either be provided
or can be easily found in your textbook or the internet.
The problem gives the mass of the water,
and the first delta t, or change in temperature, can be seen on the graph.
it is 20 °C, and the second delta t is 50 °C.
Let’s bring back the two equations for reference.
There are three enthalpy changes
and we now have all the values needed to plug into the equations.
The first enthalpy change is the mass, 10 grams,
times the specific heat of ice, 2.1 J/g°C,
times the temperature change, 20°C.
Plug in the numbers for delta H one.
You can see that grams and Celsius cancel,
leaving joules, the correct unit for heat energy and for enthalpy.
For the second energy change,
we have the heat of fusion and the mass.
But we need moles since heat of fusion is given per mole,
and so the molar mass of water is used to convert mass to moles.
Also, since delta H one is in joules,
we convert kilojoules to joules to keep the units consistent:
…grams and mols cancel, leaving joules for delta H two.
Finally, the enthalpy change of heating the liquid water, delta H three.
We plug in the correct values.
Water boils at 100 °C but we are only going to 50 °C,
so that is the last calculation.
The total delta H is the three heat changes added together,
which gives 5852 J of energy.
In other words, 5852 J of heat energy is required
to change the temperature of 10 g of water from -20 to +50 °C.
Let’s try another problem: What is the total enthalpy change for 10 g of paradichlorobenzene when going from 200 °C to 80 °C?
We are no longer dealing with water, and so we have a completely different set of values for the specific heats, for the heats of phase changes, and for the melting and boiling temperature.
Using these values, we will work out each delta H. You may want to turn off the video to first see if you can do the calculations yourself.
First let’s draw a graph to determine where each delta H is occurring. In this problem the temperature is decreasing, and so now heat will be lost instead of gained, but the calculations are exactly the same as before, which you will see.
The first thing is to check to see where phase changes occur. We see from the data provided that boiling occurs at 123 °C.
Our starting temperature is above that,
and so the PDCB is starting out as a gas.
As heat energy is lost and the temperature decreases to 123 °C,
the gas will condense to liquid,
keeping in mind that the boiling point
is the same as the condensation point.
So there will be a phase change at 123 °C from gas to liquid PDCB.
We see that the next phase change occurs below 80 °C which we will not reach and so that can be ignored.
Drawing the temperature versus energy change
shows the three energy changes occurring.
Let’s put up the two equations we will use and start plugging in the given values.
The first energy change occurs in the gas phase so we use the specific heat of the gas, and the temperature change is shown on the graph to be 77 °C. Grams and celsius cancel so we are left with joules, the desired unit.
For the second delta H, we’ll need heat of vaporization, which is the energy needed to boil one mole of PDCB. But here we have condensation. The magnitude of the energy change is the same, just in the opposite direction, so we can use the value given for heat of vaporization.
However, since it is given in kJ per mole, we will need the molar mass of PDCB to convert mass to moles, and to stay consistent we convert kilojoules to joules. All units cancel except joules.
For delta H 3 we use the specific heat of the liquid and the temperature change given by the graph, which is 43 °C.
The calculations added together give 1658.4 J.
But since heat was lost here when cooling from 200 to 80 °C,
the enthalpy change is negative;
It is an exothermic process.
Ten grams of PDCB loses 1658.4 joules
when decreasing in temperature from 200 to 80 °C.
Keep in mind when doing these problems
that drawing a graph to guide your calculations is very useful.
That’s it!