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Turning degrees on a thermometer into kilojoules per mole, and why the measured value always comes out short.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to take a mass of water and a temperature change, work out the energy in joules and then in kilojoules, divide by the amount in moles that reacted, and decide the sign from the direction the thermometer moved. You will also be able to rearrange the relation for any one of its quantities, and to say why a school measurement of an enthalpy change comes out smaller in size than the book value and never larger.
You can say whether a change is exothermic or endothermic and what sign goes with it, and you can turn a mass into an amount in moles. This lesson puts a number on the size of the change, using the only instrument that can measure one: a thermometer in a beaker of water.
| Term | What it means |
|---|---|
| Calorimetry | Measuring an energy change by measuring a temperature change. |
| Specific heat capacity | The energy that raises one gram by one kelvin; 4.2 J/(g·K) for water. |
| Calorimeter | Whatever holds the water; in a school, usually a foam cup with a lid. |
| Temperature change | One kelvin and one degree Celsius are the same size of change. |
| Systematic error | An error that pushes every result the same way. |
Everything in this lesson is one chain.
$$q = m \times c \times \Delta T \qquad\text{then}\qquad \Delta H = \pm\,\frac{q}{n}$$
The decision is the part to slow down on. $q = mc\Delta T$ is a statement about the water and it is positive either way, because the water gained energy in one case and lost it in the other and $m$, $c$ and $\Delta T$ are all positive numbers. Nothing in that formula knows about the reaction. You supply that at the end, from the direction the thermometer moved.
One unit note. Some answers here come with a unit selector beside the box and some with the unit simply printed. The rule is about the unit, not about the quantity: a single unit — a gram, a joule, a kilojoule — is one the school can convert, so you choose it. A unit that is one thing divided by another — kilojoules per mole, kilojoules per gram — is not, so it is printed and you supply only the number. The arithmetic is the same either way.
And one note about digits. One question in this lesson asks for its answer to three significant figures and marks them. That is not fussiness: the energy and the temperature change it is worked out from were measured to three figures, and a mass quoted to six would be claiming five digits nobody read off an instrument. Every other question in this course lets you write as many digits as you like, and the one that does not says so in the question.
Another way: picture
The water is the messenger. The reaction does something you cannot see, the water carries the news as a temperature change, and $m \times c \times \Delta T$ is how you translate the news back into energy. A big beaker of water moving one degree and a small one moving ten degrees can be carrying exactly the same message.
Another way: steps
For any calorimetry question:
Identify the water. Find the mass of the liquid whose temperature was measured. A dilute solution counts as water: 100 mL of it is 100 g.
Find the temperature change. Final minus starting, taken as a positive number of kelvin.
Find the energy. Mass times 4.2 times the change gives joules. Divide by 1000 for kilojoules.
Find the amount that reacted. If you were given a mass of fuel, divide it by the fuel's molar mass. If two solutions reacted, the amount is the amount of the one that limited.
Divide and sign. Kilojoules over moles, then a minus sign if the water warmed and a plus sign if it cooled.
Check the work. Is the energy sensible for a school experiment — a few kilojoules, not thousands? A result of 8400 kJ means the division by 1000 was missed. Does the sign agree with what the thermometer did? And compare with the book value if you have one: a school result should be smaller in size, never larger. A measured combustion enthalpy bigger than the book's points to an arithmetic slip, because every source of error in this experiment loses energy.
Using the water's mass is required because $q = mc\Delta T$ describes the substance whose temperature changed, and that is the water. The reacting substance's mass belongs to a different step.
Multiplying by the specific heat capacity is allowed because it is a rate: joules per gram per kelvin. Times grams and times kelvin, it leaves joules.
Taking $\Delta T$ as positive is allowed because the formula measures how much energy the water exchanged, not which way. Direction is a separate question, answered by the thermometer.
Dividing by moles is allowed because the energy released is proportional to the amount that reacted, so energy per mole is the same however big the experiment was.
Adding the sign by hand is required because the enthalpy change is the reaction's change, and the reaction's energy moves opposite to the water's. Water that warmed received energy from the reaction, which therefore lost it.
A foam cup with a lid is a decent calorimeter for a neutralization and a poor one for a burning fuel. In both cases some of the energy never reaches the water, and it always goes the same way — out.
Where it goes:
Every one of these makes $\Delta T$ smaller than it should be, so $q$ is smaller, so the size of $\Delta H$ comes out below the book value. That is a systematic error: it pushes the answer the same way every time, and repeating the experiment more carefully will not fix it, though insulating the cup and fitting a lid will reduce it.
This is worth knowing as a check rather than an excuse. A measured enthalpy of combustion for ethanol of about $-900$ kJ per mole against a book value of $-1367$ is a normal school result. One of $-1500$ is not too good; it is a sign of an arithmetic mistake, because the error in this experiment does not point that way.
Two quantities in this calculation are masses and they belong to different substances. Getting them the wrong way round is the commonest way to a wrong answer that looks reasonable.
| Symbol | Whose | Where it is used |
|---|---|---|
| $m$ | the water | in $q = mc\Delta T$, and nowhere else |
| $n$ | the substance that reacted | in the last division only |
The mass of fuel burned or salt dissolved never appears in $q = mc\Delta T$. It appears earlier, if at all, as the thing you turned into $n$ with a molar mass.
Mass is not amount. Two grams of hydrogen and two grams of lead are the same mass and nothing like the same number of atoms, and a reaction counts atoms. Every question in this unit that looks like it is about mass is about moles with mass at each end.
So a full question often has five steps rather than four: mass of fuel, molar mass, amount in moles, then the chain. The amount is worked out first and set aside, and the calorimetry is done without looking at it.
Every packaged food sold in the United States carries a Nutrition Facts label, required by the FDA, and the Calories on it began as calorimetry. A food Calorie, with a capital C, is a kilocalorie: about 4.2 kJ, the energy that warms a kilogram of water by one degree Celsius — the same 4.2 that appears in this lesson's formula, scaled up a thousand times.
The original measurements were made in a bomb calorimeter: a sealed steel vessel, filled with oxygen, sitting in a known mass of water. A dried food sample is ignited inside, it burns completely, and the water's temperature rise gives the energy, by $q = mc\Delta T$. Suppose 1.0 g of dried peanut warms 2000 g of water by 2.9 K. Then $q = 2000 \times 4.2 \times 2.9 = 24{,}360$ J, about 24 kJ, or 5.8 Calories per gram — close to the value for peanuts.
The sealed vessel is the point. An open beaker over a flame, like the school version, loses much of the energy to the air and reads low, for the reasons this lesson lists. A bomb calorimeter traps nearly all of it. Today most labels are worked out from standard values per gram of fat, protein and carbohydrate — 9, 4 and 4 Calories — but those standard values were themselves measured by calorimetry, more than a century ago, by the American chemist Wilbur Atwater.
Heating a liter of water from 20 °C to 100 °C takes $1000 \times 4.2 \times 80 = 336{,}000$ J, about 336 kJ. A 1500-watt kettle supplies 1.5 kJ every second, so it needs about four minutes, a little longer in practice because some energy warms the kettle itself.
Using the mass of the fuel in $q = mc\Delta T$. The formula is about the substance whose temperature changed, and that is the water. A gram of ethanol in the burner has nothing to do with it.
Forgetting to divide by a thousand. An answer of $8400$ kJ per mole for a school experiment is three orders of magnitude out and should look wrong on sight. $q$ comes out in joules; enthalpy changes are quoted in kilojoules.
Leaving out the sign, or taking it from the arithmetic. $q$ is positive whichever way the temperature went. The sign of $\Delta H$ is a separate decision made from the direction of the change.
Forgetting to divide by the amount. An energy in kilojoules is not an enthalpy change. The whole point of per mole is that it does not depend on how much you happened to use.
Expecting the measured value to match the book. It will not, and it will be short rather than over. A result that matches exactly in an open cup is more suspicious than one that is twenty percent low.
Read the measurements.
$100 \text{ g water}, \ 20 \text{ K rise}, \ 0.05 \text{ mol}$
Water's mass and change; fuel's amount.
Find the energy in joules.
$100 \times 4.2 \times 20 = 8400 \text{ J}$
Mass, capacity, change of the water.
Convert to kilojoules.
$8400 \div 1000 = 8.4 \text{ kJ}$
The unit for enthalpy.
Divide by the moles.
$8.4 \div 0.05 = 168$
Per mole of ethanol.
Sign it and compare.
$-168 \text{ against } -1367 \text{ kJ/mol}$
An open beaker loses most of the energy.
Read the mixture.
$50 \text{ mL acid} + 50 \text{ mL base} = 100 \text{ g}$
A dilute solution counts as water.
Read the temperature change.
$6 \text{ K rise}$
The water warmed.
Find the energy.
$100 \times 4.2 \times 6 = 2520 \text{ J} = 2.52 \text{ kJ}$
The usual product.
Find the amount reacted.
$0.05 \text{ mol of water formed}$
One to one, from each solution.
Divide and sign.
$-2.52 \div 0.05 = -50.4 \text{ kJ/mol}$
The temperature rose.
Compare with the book.
$-50.4 \text{ against about } -57$
A lidded cup comes much closer.
Read the measurements.
$0.32 \text{ g methanol}; 100 \text{ g water rises } 15 \text{ K}$
The fuel is given as a mass.
Find methanol's molar mass.
$12 + 4 + 16 = 32$
$\mathrm{CH_3OH}$.
Convert the fuel to moles.
$0.32 \div 32 = 0.01 \text{ mol}$
Set aside for the end.
Find the energy.
$100 \times 4.2 \times 15 = 6300 \text{ J}$
About the water only.
Convert to kilojoules.
$6.3 \text{ kJ}$
Divide by a thousand.
Divide by the moles.
$6.3 \div 0.01 = 630$
Per mole of methanol.
Sign and judge it.
$-630 \text{ against } -726 \text{ kJ/mol}$
Short, in the expected direction.
Find the energy.
$200 \times 4.2 \times 5 = 4200 \text{ J} = 4.2 \text{ kJ}$
Mass of water, capacity, temperature change.
Divide by the moles.
$4.2 \div 0.1 = 42 \text{ kJ/mol}$
Per mole of salt.
Decide the sign.
In an acid and an alkali mixed in a polystyrene cup, $100$ g of water changed temperature by $20$ K — it rose — and $0.15$ mol of hydrochloric acid was involved. What is the enthalpy change per mole? Take the capacity of water as $4.2$ J per gram per kelvin, and give the sign.
Answer: kJ/mol
Complete the worked solution: an alcohol burner warms one hundred fifty grams of water by twelve kelvin while six hundredths of a mole of fuel burns. Water's capacity is four point two joules per gram per kelvin. Find the energy in joules, then in kilojoules, then the enthalpy change per mole with its sign.
Find the energy in joules.
$\text{one fifty} \times \text{four point two} \times \text{twelve} =$ q
Mass of water, capacity, temperature change.
Convert to kilojoules.
$(\text{joules}) \div \text{one thousand} =$ j
The unit enthalpy uses.
Divide by moles and sign it.
$-(\text{kilojoules}) \div \text{six hundredths} =$ h
The water warmed, so negative.
a spirit burner of ethanol heating a beaker of water is done three times: once as written, once with twice as much of everything, and once with four times as much. The amount of ethanol is given for each. Fill in the energy and the enthalpy change per mole.
| energy released or taken in, in kJ | amount that reacted, in mol | enthalpy change, in kJ per mole | |
|---|---|---|---|
| the experiment as written | 0.05 | ||
| twice as much of everything | 0.1 | ||
| four times as much of everything | 0.2 |
A technician repeating an acid and an alkali mixed in a polystyrene cup recorded that the water gained $25200$ J and that its temperature rose by $20$ K, but forgot to write down how much water was in the beaker. What mass of water was it? Take the specific heat capacity of water as $4.2$ J per gram per kelvin, and answer in grams, to three significant figures.
Answer: unit: g / kg / mg
A drip coffee maker heats $2000$ g of water from $20$ °C to $95$ °C. How much energy, in kJ, does the water gain? Use $4.2$ J per gram per kelvin.
The answer: a kJ.
A food laboratory works out the energy in a snack by burning a dried sample of it under a beaker of water. It burns $0.7$ g of the sample under $200$ g of water, and the temperature of the water rose by $10$ K. How much energy did the sample release per gram? Take the capacity of water as $4.2$ J per gram per kelvin.
Answer: kJ/g
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
In a larger acid and alkali mixture in a polystyrene cup, the temperature of the water rose by $20$ K. The mass of water and the amount of nitric acid involved are given. Work down the table to the enthalpy change. Take the specific heat capacity of water as $4.2$ J per gram per kelvin.
| value | |
|---|---|
| mass of water, in g | 600 |
| temperature change of the water, in K | 20 |
| energy the water gained or lost, in J | |
| the same energy, in kJ | |
| amount of nitric acid that reacted, in mol | 0.9 |
| enthalpy change, in kJ per mole |
You can turn a temperature change into an enthalpy change in kilojoules per mole, and you know which mass goes into the calculation and which does not. Say what energy is gained by two hundred grams of water warming by ten kelvin, and where the sign of the answer comes from. Next: working the same number out from the bonds that broke and the bonds that formed, and why that answer is only ever an estimate.
15. Your turn: dissolving $0.1$ mol of a salt in $200$ g of water makes the temperature fall by $5$ K. What is the enthalpy change of solution?, step 3
$+42 \text{ kJ/mol}$
The temperature fell, so energy went in.