One Equation Needed

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Problems using two parts of the T-T graph

Problems using three parts of the T-T graph

Problems using four parts of the T-T graph

Problems using five parts of the T-T graph

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The following problem is the equivalent of area number 3 on the time-temperature graph. The ONLY thing happening in this problem is a temperature change. There are no phase changes at all.

**Problem #1:** Calculate the heat necessary to raise 27.0 g of water from 10.0 °C to 90.0 °C

The important factor about this problem is that a ONLY temperature change is involved. Therefore, the equation to use is:

q = (mass) (Δt) (C_{p})

This summarizes the information needed:

Δt = 80.0 °C

The mass = 27.0 g

C_{p}= 4.184 J g¯^{1}°C¯^{1}

Only one calculation is needed and it is:

q = (27.0 g) (80.0 °C) (4.184 J g¯^{1}°C¯^{1})

Multiply it out and round off to the proper number of sig figs.

By the way, how did I know to use the specific heat value for liquid water? It is because I knew the melting point of water (0 ° C) and the boiling point of water (100 ° C). The temperature range was entirely within those two values, so I know that the water stayed as a liquid the entire time.

Less often, you will see problems which involve water changing temperature as a solid (area one on the time-temperature graph) or water changing temperature as a gas (area five on the time-temperature graph). Here is an example of each:

**Problem #2:** Calculate the heat necessary to raise 27.0 g of water from -90.0 °C to -10.0 °C (graph area one)

**Problem #3:** Calculate the heat necessary to raise 27.0 g of water from 110.0 °C to 190.0 °C (graph area five)

The technique is the same as in the liquid water example (area three of the graph) above. Here are the solutions:

Basic equation:q = (mass) (Δt) (C_{p})

Problem #2:q = (27.0 g) (80.0 °C) (2.06 J g¯^{1}°C¯^{1})

Problem #3:q = (27.0 g) (80.0 °C) (2.02 J g¯^{1}°C¯^{1})

Notice that all three of the above problems are exactly the same in mass and Δt. That was deliberate. Suppose the masses were 50.0 g. There would be no difference in technique, just that 50.0 g. would be the mass instead of 27.0 g. The same idea holds true for the change in temperature.

The critical thing to notice are the different specific heats. You MUST discern from the problem if the water (or whatever substance is in the problem) is a solid, liquid or gas. Then, you must select the proper specific heat (2.06 for solid, 4.184 for liquid or 2.02 for gas). Also, please note that your reference source may provide different values than the ones I am using.

To select the proper specific heat, you must know the melting and boiling temperature for the substance. Since water is the most-commonly discussed substance, I recommend you know the melting/freezing temperature for water (0 °C) and the boiling/condensing temperatue (100 °C) for water. Teachers (and textbook question writers) often assume you already know those two values.

What you would do is look at the temperatures in your problem. Do they fall below 0 (solid), between 0 to 100 (liquid) or above 100 (gas)? Also, if your starting or ending temperatures cross 0 or 100, then you know you have a problem that requires more than the one type of calculation discussed in this tutorial.

Please be aware that another type of "one equation required" problem might look like these two:

**Problem #4:** How much energy is required to completely boil away 100.0 g of water at 100.0 °C? (area four on the graph)

**Problem #5:** How much energy is required to melt 100.0 g of water at 0 °C? (area two on the graph)

The first problem requires the use of the molar heat of vaporization and the second requires the use of the molar heat of fusion.

Here are the two solutions:

40.7 kJ/mol x (100.0 g / 18.0 g/mol)6.02 kJ/mol x (100.0 g / 18.0 g/mol)

Often these problems are solved using the heat of vaporization (2259 J/g) or the heat of fusion (334.166 J/g). These terms do not use the word molar. Here are the solutions:

2259 J/g x 100.0 g334.166 J/g x 100.0 g

The heat of vaporization comes from 40700 J / 18.015 g and the heat of fusion comes from 6020 J / 18.015 g.

**Problem #6:** Calculate the heat given off when 159.7 g of copper cools from 155.0 °C to 23.0 °C. The specific heat capacity of copper is 0.385 J/g °C.

**Solution:**

q = (159.7 g) (132.0 °C) (0.385 J/g °C)q = 8115.954 J = 8.12 kJ (to three sig figs)

**Problem #7:** How many joules of energy would be required to heat 15.9 g of diamond from 23.6 °C to 54.2 °C? (Specific heat capacity of diamond = 0.5091 J/g °C.)

**Solution:**

q = (15.9 g) (30.6 °C) (0.5091 J/g °C)q = 247.697514 J = 248 J (to three sig figs)

This problem show a different twist to the "one equation" problems shown above:

**Problem #8:** Assume that 491.8 J of heat is added to 5.00 g of water originally at 23.0 °C. What would be the final temperature of the water? (Specific heat capacity of water = 4.184 J/g °C.)

**Solution:**

491.8 J = (5.00 g) (x) (4.184 J/g °C)x = 23.5 °C

However, 23.5 is the change in temperature. To get the final temperature, we do this:

23.0 °C + 23.5 °C = 46.5 °C

Go to the Time-Temperature Graph tutorial

Problems using two parts of the T-T graph

Problems using three parts of the T-T graph

Problems using four parts of the T-T graph