Learn Calorimetry & enthalpy from scratch
Before this: Stoichiometry
1Heat and temperature
Heat (q) is energy that flows from a hot object to a colder one. It is measured in joules (J) or kilojoules (1 kJ = 1000 J).
Temperature tells you how hot something is. Heat is what changes the temperature.
Heating 1 kg of water by 10 °C takes more energy than heating 1 kg of iron by 10 °C. Each substance has its own specific heat c: the heat needed to warm 1 g by 1 °C.
For water, c = 4.18 J/(g·°C). For aluminium it is 0.897 J/(g·°C), for lead only 0.129 J/(g·°C). Water is hard to heat up, metals are easy.
The heat needed for a temperature change is:
q = m · c · ΔT
Here m is the mass in grams and ΔT = Tfinal − Tinitial. A positive q means heat is taken in; a negative q means heat is given off.
Self-check How much heat is needed to warm 100 g of water by 10 °C (c = 4.18 J/(g·°C))?
- 0.418 kJ
- 4.18 kJ
- 41.8 kJ
- 418 J
Answer: 4.18 kJ. q = 100 · 4.18 · 10 = 4180 J = 4.18 kJ. 418 J forgets to multiply by ΔT, and 41.8 kJ slips a factor of 10 when converting J to kJ.
2The calorimeter: heat lost = heat gained
A calorimeter is an insulated cup. No heat leaves it, so all heat given off by one thing is taken in by the other.
Drop a hot metal block into cold water. The block cools down, the water warms up, until both reach the same final temperature Tf.
mm · cm · (Tm − Tf) = mw · cw · (Tf − Tw)
The left side is the heat lost by the metal, the right side the heat gained by the water. Both are written as positive numbers.
Solving for Tf gives a weighted average of the two start temperatures, where each weight is m · c:
Tf = (mmcmTm + mwcwTw) / (mmcm + mwcw)
Because water has a large m · c, Tf usually ends up close to the water temperature.
Self-check A hot metal block is dropped into cold water in an insulated cup. Where must the final temperature lie?
- Above the metal’s start temperature
- Below the water’s start temperature
- Between the two start temperatures
- Exactly halfway between the two start temperatures
Answer: Between the two start temperatures. The metal cools and the water warms until they meet, so Tf lies between them. It is halfway only if both have the same m · c, which is rare.
3Enthalpy of reaction
Reactions also take in or give off heat. At constant pressure this heat is the enthalpy change ΔH, in kJ.
- Exothermic: ΔH is negative. The reaction gives off heat, like burning fuel.
- Endothermic: ΔH is positive. The reaction takes in heat, like a cold pack.
The standard enthalpy of formation ΔH°f is the ΔH for making 1 mol of a substance from its elements. For an element in its normal form, such as O₂ or C, it is 0.
With these values from a table, you find ΔH° of any reaction:
ΔH° = ΣΔH°f(products) − ΣΔH°f(reactants)
Multiply each ΔH°f by the coefficient of that substance in the balanced equation. This works because of Hess’s law: the total ΔH does not depend on the route from start to end.
Self-check Find ΔH° for CH₄ + 2O₂ → CO₂ + 2H₂O(l). Use ΔH°f: CH₄ −74.8, CO₂ −393.5, H₂O(l) −285.8 kJ/mol.
- +890.3 kJ
- −1039.9 kJ
- −604.5 kJ
- −890.3 kJ
Answer: −890.3 kJ. Products: −393.5 + 2 · (−285.8) = −965.1. Minus reactants: −965.1 − (−74.8) = −890.3 kJ. −604.5 forgets the coefficient 2 of water, and +890.3 subtracts the wrong way round.
4The method step by step
Heating:
- Write q = m · c · ΔT and fill in what is given. Convert kJ to J (× 1000) so the units match c.
- Solve for the unknown: q, ΔT, m or c.
Calorimeter:
- Write heat lost by the hot object = heat gained by the cold one.
- Put Tf in both temperature differences, hot minus Tf and Tf minus cold.
- Solve for Tf, and check that it lies between the two start temperatures.
Enthalpy of formation:
- Multiply every ΔH°f by its coefficient. Elements count as 0.
- ΔH° = sum for the products − sum for the reactants.
- For the heat from a given mass, convert the mass to moles and scale ΔH° to that amount.
5A worked example
Follow every step. Try to predict the next one before you open it.
Worked exampleA 100 g block of aluminium (c = 0.897 J/(g·°C)) at 110 °C is dropped into 200 g of water (c = 4.18 J/(g·°C)) at 20 °C in an insulated cup. Find the final temperature Tf.
1Aluminium: 100 · 0.897 = 89.7 J/°C. Water: 200 · 4.18 = 836 J/°C.
The water needs far more heat per degree, so Tf will be close to 20 °C.
2Heat lost by the aluminium = heat gained by the water.
89.7 · (110 − Tf) = 836 · (Tf − 20)
39867 − 89.7·Tf = 836·Tf − 16 720
Tf = (9867 + 16 720) / (89.7 + 836) = 26 587 / 925.7 = 28.7 °C
4Heat lost by the aluminium: 89.7 · (110 − 28.7) ≈ 7290 J. Heat gained by the water: 836 · (28.7 − 20) ≈ 7290 J. They are equal.
And 28.7 °C lies between 20 °C and 110 °C, close to the water temperature, as expected.
6Common mistakes
- Mixing J and kJ. Specific heats are in J, so convert kJ to J first.
- Using the wrong mass, for example the mass of the metal with the specific heat of water.
- Writing ΔT as initial minus final. Then the sign of q is wrong.
- In a calorimeter, writing both sides as Tf minus start. Then one side is negative and the balance fails.
- Forgetting the coefficients when adding enthalpies of formation.
- Calculating reactants minus products, which flips the sign of ΔH°.
Self-check 78 kJ of heat is added to 400 g of water (c = 4.18 J/(g·°C)). What is the temperature rise?
- 0.0214 °C
- 0.0467 °C
- 195 °C
- 46.7 °C
Answer: 46.7 °C. ΔT = q / (m · c) = 78 000 / (400 · 4.18) = 46.7 °C. Leaving q in kJ gives 0.0467 °C, forgetting c gives 78 000 / 400 = 195 °C, and turning the fraction upside down gives 0.0214 °C.
7How exam questions look
- Easy: heating with q = m·c·ΔT. Find the heat to warm water between two temperatures, or the temperature rise from a given heat.
- Medium: calorimeter. A hot metal block in cold water; find the final temperature with heat lost = heat gained.
- Hard: enthalpy of formation. Find ΔH° of a combustion from a table, or the heat released when a given mass burns.
Example of a hard question: ethanol burns with ΔH° = −1366.7 kJ/mol. 50 g of ethanol (M = 46.07 g/mol) is 1.085 mol, so it releases 1.085 · 1366.7 ≈ 1483 kJ.
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