The Time-Temperature Graph


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We are going to heat a container that has 72.0 grams of ice (no liquid water yet!) in it. To make the illustration simple, please consider that 100% of the heat applied goes into the water. There is no loss of heat into heating the container and no heat is lost to the air.

Let us suppose the ice starts at minus 10.0 °C and that the pressure is always one atmosphere. We will end the example with steam at 120.0 °C.

There are five major steps to discuss in turn before this problem is completely solved. Here they are:

1) the ice rises in temperature from -10.0 to 0.00 °C.
2) the ice melts at 0.00 °C.
3) the liquid water then rises in temperature from zero to 100.0 °C.
4) the liquid water then boils at 100.0 °C.
5) the steam then rises in temperature from 100.0 to 120.0 °C

Each one of these steps will have a calculation associated with it. WARNING: many homework and test questions can be written which use less than the five steps. For example, suppose the water in the problem above started at 10.0 °C. Then, only steps 3, 4, and 5 would be required for solution.

 

To the right is the type of graph which is typically used to show this process over time.

The ChemTeam hopes that you can figure out that the five numbered sections on the graph relate to the five numbered parts of the list just above the graph.

Also, note that numbers 2 and 4 are phases changes: solid to liquid in #2 and liquid to gas in #4.

Here are some symbols that will be used, A LOT!!

1) Δt = the change in temperature from start to finish in degrees Celsius (°C)
2) m = mass of substance in grams
3) Cp = the specific heat. Its unit is Joules per gram-degree Celsius (J / g °C is one way to write the unit; J g¯1 °C¯1 is another)
4) q = the amount of heat involved, measured in Joules or kilojoules (symbols = J and kJ)
5) mol = moles of substance.
6) ΔHfus is the symbol for the molar heat of fusion and ΔHvap is the symbol for the molar heat of vaporization.
We will also require the molar mass of the substance. In this example it is water, so the molar mass is 18.0 g/mol.

By the way, the p means the specific heat is measured at constant pressure; there is a related specific heat we will not discuss (yet) which is measured at constant volume. Not too suprisingly (I hope), it has the symbol Cv.


Step One: solid ice rises in temperature

As we apply heat, the ice will rise in temperature until it arrives at its normal melting point of zero Celsius.

Once it arrives at zero, the Δt equals 10.0 °C.

Here is an important point: THE ICE HAS NOT MELTED YET.

At the end of this step we have SOLID ice at zero degrees. It has not melted yet. That's an important point.

Each gram of water requires a constant amount of energy to go up each degree Celsius. This amount of energy is called specific heat and has the symbol Cp.

Follow this link to calculate the energy used in this step.


Step Two: solid ice melts

Now, we continue to add energy and the ice begins to melt.

However, the temperature DOES NOT CHANGE. It remains at zero during the time the ice melts.

Each mole of water will require a constant amount of energy to melt. That amount is named the molar heat of fusion and its symbol is ΔHfus. The molar heat of fusion is the energy required to melt one mole of a substance at its normal melting point. One mole of solid water, one mole of solid benzene, one mole of solid lead. It does not matter. Each substance has its own value.

During this time, the energy is being used to overcome water molecules' attraction for each other, destroying the three-dimensional structure of the ice.

The unit for this is kJ/mol. Sometimes you see older references that use kcal/mol. The conversion between calories and Joules is 4.184 J = 1.000 cal.

Sometimes you also see this number expressed "per gram" rather than "per mole." For example, water's molar heat of fusion is 6.02 kJ/mol. Expressed per gram, it is 334.16 J/g.

Notice how I shifted to Joules instead of kilojoules. This was done to keep the number within the ranges of ones to hundreds. Writing the value using kJ would require I write 0.33416. It is more understandable to write 334.16.

Typically, the term "heat of fusion" is used with the "per gram" value.

Follow this link to calculate the energy used in this step.


Step Three: liquid water rises in temperature

Once the ice is totally melted, the temperature can now begin to rise again.

It continues to go up until it reaches its normal boiling point of 100.0 °C.

Since the temperature went from zero to 100, the Δt is 100.

Here is an important point: THE LIQUID HAS NOT BOILED YET.

At the end of this step we have liquid water at 100 degrees. It has not turned to steam yet.

Each gram of water requires a constant amount of energy to go up each degree Celsius. This amount of energy is called specific heat and has the symbol Cp. There will be a different value needed, depending on the substance being in the solid, liquid or gas phase.

Follow this link to calculate the energy used in this step.


Step Four: liquid water boils

Now, we continue to add energy and the water begins to boil.

However, the temperature DOES NOT CHANGE. It remains at 100 during the time the water boils.

Each mole of water will require a constant amount of energy to boil. That amount is named the molar heat of vaporization and its symbol is ΔHvap. The molar heat of vaporization is the energy required to boil one mole of a substance at its normal boiling point. One mole of liquid water, one mole of liquid benzene, one mole of liquid lead. It does not matter. Each substance has its own value.

During this time, the energy is being used to overcome water molecules' attraction for each other, allowing them to move from close together (liquid) to quite far apart (the gas state).

The unit for this is kJ/mol. Sometimes you see older references that use kcal/mol. The conversion between calories and Joules is 4.184 J = 1.000 cal.

Sometimes you also see this number expressed "per gram" rather than "per mole." For example, water's molar heat of vaporization is 40.7 kJ/mol. Expressed per gram, it is 2261 J/g or 2.26 kJ/g.

Typically, the term "heat of vaporization" is used with the "per gram" value.

Follow this link to calculate the energy used in this step.


Step Five: steam rises in temperature

Once the water is completely changed to steam, the temperature can now begin to rise again.

It continues to go up until we stop adding energy. In this case, let the temperature rise to 120 °C.

Since the temperature went from 100 to 120, the Δt is 20.

Each gram of water requires a constant amount of energy to go up each degree Celsius. This amount of energy is called specific heat and has the symbol Cp. There will be a different value needed, depending on the substance being in the solid, liquid or gas phase.

Follow this link to calculate the energy used in this step.

Follow this link for the final summing up and the answer.


More example problems

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