Calorimetry — measuring enthalpy experimentally — Chemistry

Before this

This leads to

A thermometer reading that is waiting to become a number

In the first lecture of this topic, Amira mixed sodium hydroxide solution with hydrochloric acid in a polystyrene cup. The temperature rose by 6.2 °C. We could say then that the reaction was exothermic and that $Δ H$ was negative. That lecture promised that this lecture would turn the 6.2 °C into a value of $Δ H$ .

Here is the difficulty. A thermometer measures temperature, in °C. An enthalpy change is energy, in kJ/mol. Those are different quantities. Every $Δ H$ value in the data tables of the last four lectures was measured by someone who crossed that gap. The method is called calorimetry. This lecture shows how it works, and why Amira's answer will come out about 10% too small.

Heat and temperature are not the same thing

Put the same small flame under a cup of water and under a large pan of water for one minute. Both receive about the same heat. The cup gets much hotter. So the temperature rise depends on how much heat goes in, and also on how much material shares it.

It also depends on the material. One gram of water needs about 4.18 J of heat to warm by 1 °C. One gram of copper needs only about 0.39 J. This amount, for 1 g and 1 °C, is called the specific heat capacity, $c$ . Water's value is large, which makes water a good place to collect heat and measure it.

Heat taken in by a material

$q = m c Δ T$

$q$ is the heat in J, $m$ the mass in g, $c$ the specific heat capacity in J/g/°C, and $Δ T$ the change in temperature in °C. A change of 1 °C is the same size as a change of 1 K, so $c$ is often written in J g⁻¹ K⁻¹.

Each part of the formula has a reason. Twice the mass needs twice the heat for the same rise. Twice the rise needs twice the heat. And $c$ converts "grams times degrees" into joules for that material. So 100 g of water warmed by 1 °C takes in $100 \times 4.18 \times 1 = 418$ J.

From a temperature rise to ΔH

Recall from the first lecture of this topic: the reacting chemicals are the system, and the water around them is the surroundings, together with the thermometer. In a calorimeter the reaction happens in the water. So the water is where the energy shows up. If the water warms up, the reaction gave out heat. If the water cools down, the reaction took heat in.

That gives a chain of three steps:

Find the heat taken in by the water: $q = m c Δ T$ .
Find the amount, in moles, of the substance that the $Δ H$ is defined for.
Divide, change J to kJ, and give the sign from the direction of the temperature change.

ΔH from a calorimeter

$Δ H = - \frac{q}{n}$ , where $q = m c Δ T$ for the solution and $n$ is the amount in moles, per mole "as defined" (for example, per mole of water formed).

Temperature rises: the surroundings gained heat, so $Δ H$ is negative. Temperature falls: the surroundings lost heat, so $Δ H$ is positive.

The minus sign is not a rule to learn. It says that the heat the water gained is heat the reaction lost.

Worked Example 1 — Amira's 6.2 °C

Suppose Amira mixed 50.0 cm³ of 1.00 mol/dm³ hydrochloric acid with 50.0 cm³ of 1.00 mol/dm³ sodium hydroxide solution. The temperature rose by 6.2 °C. Find the enthalpy change of neutralisation, $Δ H_{neut}$ , per mole of water formed.

HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

Mass: 100.0 cm³ of dilute solution. Assume it has the density of water, 1.00 g/cm³, so $m = 100.0$ g.

Heat: $q = 100.0 \times 4.18 \times 6.2 \approx 2592$ J $\approx 2.59$ kJ.

Moles: $n (HCl) = 1.00 \times \frac{50.0}{1000} = 0.0500$ mol, and the same for NaOH. The ratio is 1 : 1, so 0.0500 mol of water forms.

$Δ H_{neut} = - \frac{2.59}{0.0500} \approx - 51.8$ kJ/mol.

Check: the sign is negative because the solution warmed up. ✓ But measured values are about −57 to −58 kJ/mol. Amira's answer is about 10% too small. The rest of this lecture finds the missing heat.

The polystyrene cup method

A simple calorimeter. Two polystyrene cups, one inside the other, with a lid, stand in a beaker so they do not tip. The thermometer and stirrer pass through the lid.

Polystyrene is full of trapped air, so it lets heat through very slowly. It also has a very small heat capacity, so the cup itself takes very little of the heat. A second cup and a lid slow the heat loss further.

1. Measure each solution with a pipette or burette, and put the first one in the cup.
2. Read the temperature every 30 seconds for a few minutes, to get a steady starting temperature.
3. Add the second solution at a recorded time, stir, and keep reading every 30 seconds for several more minutes.
4. Plot temperature against time, and find $Δ T$ from the graph.

Step 4 is where the method gets clever. The next section explains why a graph is better than one highest reading.

Where the missing heat went

This is the hard part of the lecture. Spend time here.

The calculation assumes that every joule from the reaction stays in the solution. Two things break that assumption. First, the solution is warmer than the room, so it starts losing heat through the cup at once. The reaction takes a few seconds, and while the temperature is still rising, heat is already leaking out. So the highest reading is lower than it would be with perfect insulation. Second, the cup, the thermometer and the stirrer warm up too. They take some of the heat, but the formula only counts the solution.

Both errors make $Δ T$ too small, so $∣Δ H ∣$ comes out too small. That is the second promise from the first lecture of this topic: some heat always escapes, and it is the main source of error in calorimetry.

The first error can be largely corrected with the graph. After the peak, the solution cools at a steady, slow rate. Draw a straight line through those cooling points and continue it back to the time of mixing. That line shows the temperature the solution would have reached if all the heat had been released at once, with no time to escape. This is called extrapolation. Try it below.

Graph builder — finding ΔT from a cooling curve

The same neutralisation (50.0 cm³ + 50.0 cm³, 1.00 mol/dm³) is done in three different containers. Readings are every 30 seconds, and the solutions are mixed at 3.0 minutes. Choose a container, then tick the box to extrapolate. (An invented model of heat loss.)

Container:

Extrapolate the cooling line back to the time of mixing

The same reaction in three containers, from an invented model. The starting temperature is 18.5 °C, and a perfect calorimeter would give a rise of about 6.8 °C (−57 kJ/mol).

Container	Highest reading	Extrapolated
Two cups, lid	25.2 °C (−56.0)	25.28 °C (−56.7)
One cup	25.1 °C (−55.2)	25.26 °C (−56.5)
Glass beaker	23.9 °C (−45.1)	24.14 °C (−47.2)

Values in brackets are ΔH in kJ/mol. Extrapolation recovers most of the heat lost through the sides. It cannot recover the heat taken by the glass.

What should you have seen? With polystyrene, the extrapolated rise was close to the full 6.8 °C. With glass, it was still far too small. The glass beaker has a large heat capacity, and the heat it took never showed on the thermometer. Extrapolation corrects for heat that leaks away over time. It does not correct for heat taken by the container. For that, the container's own heat capacity must be measured, which Try it yourself question 4 does.

Stop and think

A student says: "Our answer was 10% too small, so we need a more precise thermometer." Will a thermometer that reads to 0.01 °C fix the problem?

Show the answer

No. The thermometer read the temperature of the solution correctly. The problem is that the solution really was cooler than it should have been, because heat had escaped and the cup had taken some. A more precise reading of the wrong temperature is still wrong. This is the difference between a random uncertainty, which more precise instruments reduce, and a systematic error, which pushes every result the same way. Heat loss is systematic: it always makes $∣Δ H ∣$ too small. Better insulation and extrapolation reduce it. A better thermometer does not.

Worked Example 2 — Using the extrapolated rise

In one polystyrene cup, the same reaction starts at 18.5 °C. The highest reading is 25.1 °C. The cooling line, extrapolated back to the time of mixing, gives 25.26 °C. Find $Δ H_{neut}$ both ways.

Highest reading: $Δ T = 6.6$ °C, so $q = 100.0 \times 4.18 \times 6.6 \approx 2759$ J, and $Δ H = - 2.759 \div 0.0500 \approx - 55.2$ kJ/mol.

Extrapolated: $Δ T = 6.76$ °C, so $q \approx 2826$ J, and $Δ H \approx - 56.5$ kJ/mol.

Check: both are negative and smaller in size than −57 kJ/mol, as heat loss predicts. The extrapolated value is closer. ✓

Burning a fuel: a much harder measurement

To measure an enthalpy change of combustion, a fuel burns in a small spirit lamp under a metal can of water. The lamp is weighed before and after, to find the mass burned.

Worked Example 3 — Ethanol in a spirit lamp

Burning 0.65 g of ethanol, C₂H₅OH, raises the temperature of 200 g of water by 9.5 °C. Find $Δ H_{c}$ , and compare it with the data value, −1367 kJ/mol.

Heat: $q = 200 \times 4.18 \times 9.5 = 7942$ J $\approx 7.94$ kJ.

Moles: $M = 2 \times 12 + 6 \times 1 + 16 = 46$ g/mol, so $n = 0.65 \div 46 \approx 0.0141$ mol.

$Δ H_{c} \approx - 7.94 \div 0.0141 \approx - 562$ kJ/mol.

Check: $562 \div 1367 \approx 0.41$ . The experiment caught only about 41% of the heat. ✓ The sign is right, and the size is far too small.

Why so much worse than the cup? The flame heats the air around the can as much as the can. Hot gases rise past the sides. The can itself warms up. Some ethanol may burn incompletely, leaving soot and releasing less energy. So this method is useful for comparing fuels with the same apparatus, not for accurate values. A draught shield, a copper can close to the flame, and a lid all help.

Accurate combustion values come from a bomb calorimeter. The sample burns in pure oxygen inside a sealed steel container, which sits in a measured mass of water. The whole instrument is first calibrated by burning a substance whose energy is known exactly, so the heat taken by the steel is measured, not ignored.

When the temperature falls

Worked Example 4 — A cold pack

Ravi dissolves 10.0 g of ammonium nitrate, NH₄NO₃, in 100.0 cm³ of water. The temperature falls from 21.0 °C to 14.8 °C. Find the enthalpy change of solution per mole of NH₄NO₃.

The water cooled, so it lost heat to the dissolving process. The process is endothermic, and $Δ H$ will be positive.

Mass: everything that changed temperature is the solution, water plus solid: $m = 100.0 + 10.0 = 110.0$ g. Heat lost by the solution: $q = 110.0 \times 4.18 \times 6.2 \approx 2851$ J $\approx 2.85$ kJ.

Moles: $M = 14 + 4 + 14 + 48 = 80$ g/mol, so $n = 10.0 \div 80 = 0.125$ mol.

$Δ H = + \frac{2.85}{0.125} \approx + 22.8$ kJ/mol.

Check: data tables give about +25.7 kJ/mol (LibreTexts). The result is positive and a little too small, because heat flowed in from the warmer room and made the fall in temperature smaller. Heat exchange with the room always makes $∣Δ H ∣$ too small, whichever way it flows. ✓

Which mass should you use, 110.0 g or 100.0 g? The whole solution changes temperature, so 110.0 g describes it better. But 4.18 J/g/°C is the value for water, not for this solution. Many textbooks and mark schemes use the mass of water alone, here 100.0 g, which gives about +20.7 kJ/mol. Both are approximations. Use the mass your question or your syllabus asks for, and say which you used.

Now sort the possible errors by what they do to the result.

Sort — what does each problem do to |ΔH|?

Each card is a problem in a calorimetry experiment. Put it in the box for its effect on the size of the measured ΔH. Read the reason after each choice.

Each problem, with its effect on the size of the measured ΔH.

Too small: heat escaping through the cup; the cup and thermometer taking heat; reading the temperature before it peaks; ethanol evaporating from the wick but counted as burned; soot from incomplete combustion.
Too large: using 4.18 J/g/°C when the solution's real value is a little lower.
No effect: a thermometer that reads 0.5 °C too high at every temperature; one reactant in excess, when you divide by the moles of the one that runs out.

Three classic mistakes

Trap 1 — Using the mass of the solid in q = mcΔT

In $q = m c Δ T$ , $m$ is the mass of the solution that changed temperature, not the mass of the substance that reacted. For a fuel, it is the mass of water in the can, not the mass of fuel burned. The mass of the reactant belongs in step 2, to find the moles.

Trap 2 — Dividing by the wrong amount

Divide by the moles of the substance the $Δ H$ is defined for, and only as much as actually reacted. For neutralisation, that is the moles of water formed, set by the reactant that runs out.

Trap 3 — Forgetting the sign or the kilo

$q = m c Δ T$ gives joules. Divide by 1000 for kJ before dividing by moles, or the answer is 1000 times too big. Then give the sign from the temperature: a rise means a negative $Δ H$ .

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

Excess zinc powder is added to 50.0 cm³ of 0.200 mol/dm³ copper(II) sulfate solution. The temperature rises by 10.3 °C. Find $Δ H$ per mole of CuSO₄ for the reaction Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s).

Hint 1
The zinc is in excess, so the copper(II) sulfate decides how much reacts.

Hint 2
$m = 50.0$ g. n(CuSO₄) = $0.200 \times 50.0 \div 1000$ .

Full answer
$q = 50.0 \times 4.18 \times 10.3 \approx 2153$ J $\approx 2.15$ kJ. $n = 0.200 \times 0.0500 = 0.0100$ mol. $Δ H \approx - 2.153 \div 0.0100 \approx - 215$ kJ/mol. Negative, because the temperature rose. Check the units: kJ ÷ mol gives kJ/mol. ✓ The mass of zinc is not used at all: it is in excess, and it is not the solution.
Predict the temperature rise when 25.0 cm³ of 2.00 mol/dm³ HCl is mixed with 25.0 cm³ of 2.00 mol/dm³ NaOH, taking $Δ H_{neut} = - 57$ kJ/mol and no heat loss. Compare with 50.0 cm³ + 50.0 cm³ of 1.00 mol/dm³.

Hint 1
Work backwards: moles of water, then $q$ , then $Δ T = q \div (m c)$ .

Hint 2
$n = 2.00 \times 0.0250 = 0.0500$ mol, and $m = 50.0$ g.

Full answer
$q = 0.0500 \times 57000 = 2850$ J. $Δ T = 2850 \div (50.0 \times 4.18) \approx 13.6$ °C. With 50.0 + 50.0 cm³ of 1.00 mol/dm³, the same 0.0500 mol of water forms, but the heat is shared by 100.0 g, so $Δ T \approx 6.8$ °C. Same heat, half the mass, twice the rise. Check: $13.6 \div 6.8 = 2$ . ✓ A larger rise is easier to measure precisely, but it also makes heat loss faster.
An improved apparatus, with a copper can and a draught shield, is used for ethanol. Burning 0.92 g of ethanol raises 250 g of water by 18.0 °C. Find $Δ H_{c}$ , the percentage of the data value (−1367 kJ/mol) that was caught, and two reasons it is still too small.

Hint 1
$n = 0.92 \div 46$ .

Hint 2
$q = 250 \times 4.18 \times 18.0$ .

Full answer
$q = 250 \times 4.18 \times 18.0 = 18810$ J $\approx 18.8$ kJ. $n = 0.92 \div 46 = 0.0200$ mol. $Δ H_{c} \approx - 18.81 \div 0.0200 \approx - 941$ kJ/mol. That is about 69% of the data value, much better than the 41% of Worked Example 3. Reasons it is still small: hot gases still escape past the can; the copper can takes heat that is not counted; some combustion may be incomplete. Check: $941 \div 1367 \approx 0.69$ . ✓
Amira wants to know how much heat her cup takes. In the cup, which starts at 20.0 °C, she mixes 50.0 g of water at 40.0 °C with 50.0 g of water at 20.0 °C. The mixture reaches 29.2 °C, not 30.0 °C. (a) Find the heat capacity of the cup, in J/°C. (b) Use it to correct her 6.2 °C result from Worked Example 1.

Hint 1
The hot water lost heat. The cold water and the cup gained it. The difference between the hot water's loss and the cold water's gain went into the cup.

Hint 2
The hot water cooled by 10.8 °C, and the cold water and the cup warmed by 9.2 °C.

Full answer
(a) Heat lost by the hot water: $50.0 \times 4.18 \times 10.8 \approx 2257$ J. Heat gained by the cold water: $50.0 \times 4.18 \times 9.2 \approx 1923$ J. The cup gained the difference, about 334 J, for a rise of 9.2 °C. So its heat capacity is about $334 \div 9.2 \approx 36$ J/°C. (b) The heat released in her reaction warmed the solution and the cup: $q \approx (418 + 36) \times 6.2 \approx 2817$ J. Then $Δ H_{neut} \approx - 2.817 \div 0.0500 \approx - 56.3$ kJ/mol, much closer to −57. Check: the correction raises $∣Δ H ∣$ , as it must, because uncounted heat was found. ✓ This is how a bomb calorimeter is calibrated, on a larger scale.

The bigger idea

Every number in a data book is a measurement, and every measurement carries the method that made it. A bomb calorimeter is sealed, so it measures the energy change at constant volume, not at constant pressure. A small correction then turns that into $Δ H$ , the heat at constant pressure that the first lecture of this topic defined. The same idea, measuring heat flow precisely, is used today in a method called differential scanning calorimetry. It measures the heat taken in as a protein unfolds, which helps scientists judge how stable a medicine will be.

The topic now ends where it began. Enthalpy tells you how much heat a reaction gives out or takes in. It does not, on its own, tell you whether the reaction will happen at all: ammonium nitrate dissolves even though it takes heat in. The next topic, entropy, explains why.

Key concepts in this lesson

The heat taken in by a material is $q = m c Δ T$ , where $m$ is the mass of the solution that changed temperature and $c$ is its specific heat capacity, 4.18 J/g/°C for water.
An enthalpy change is found from $Δ H = - q / n$ , where $n$ is the amount, in moles, of the substance the $Δ H$ is defined for.
A temperature rise means the reaction released heat, so $Δ H$ is negative, and a temperature fall means $Δ H$ is positive.
Heat lost to or gained from the room, and heat taken by the container, are systematic errors that make the measured $∣Δ H ∣$ too small.
Extrapolating the cooling line back to the time of mixing corrects for heat lost during the reaction, and measuring the container's heat capacity corrects for the heat it takes.