Heat, temperature, and energy
Thermochemistry is the study of heat flow into and out of chemical reactions and physical changes. Three words get mixed up constantly, so pin them down first: - Heat (q) is energy that moves between two things because they're at different temperatures. It's a transfer, not a substance — you can't have "a heat," only "heat flowing." - Temperature is a measure of the *average kinetic energy* of the particles in a substance. It tells you how hot something is, not how much energy it holds. - Energy is the capacity to do work or transfer heat. Heat is one way energy moves; it's measured in joules (J) or kilojoules (kJ), same as any other energy. A cup of water and a swimming pool can be at the same temperature, but the pool holds far more total thermal energy because it has so much more mass. Temperature is intensive (doesn't depend on amount); total thermal energy is extensive (it does).
System vs. surroundings
Every thermochemistry problem splits the universe into the system (the reaction or substance you're tracking) and the surroundings (everything else — the beaker, the water bath, the air in the room). Heat that leaves the system enters the surroundings, and vice versa. Every sign convention below is written from the system's point of view.
Exothermic vs. endothermic
Reading ΔH
ΔH is the enthalpy change of a reaction: a negative ΔH means the system releases heat (exothermic), and a positive ΔH means it absorbs heat (endothermic). The section Enthalpy (ΔH) and sign conventions covers this in full; the one-line version is all you need for the examples just below.
Every chemical or physical change either releases heat to the surroundings or absorbs heat from them. Exothermic ("exo-" = out) reactions release heat. The system loses energy, so the surroundings get warmer. Combustion is the classic example: CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = −890 kJ/mol Burning methane releases 890 kJ of heat per mole into the surroundings — that's why a stove burner feels hot. Endothermic ("endo-" = in) reactions absorb heat. The system gains energy, so the surroundings get colder. Photosynthesis is the classic example: 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂, ΔH = +2803 kJ/mol A plant absorbs light energy to drive this reaction forward — without an outside energy source, it doesn't happen.
| Everyday example | Type | What you feel |
|---|---|---|
| Combustion (burning fuel) | Exothermic | Gets hot |
| Neutralization (acid + base) | Exothermic | Container warms slightly |
| Dissolving CaCl₂ in water (hot pack) | Exothermic | Pack warms |
| Dissolving NH₄NO₃ in water (cold pack) | Endothermic | Pack cools |
| Melting ice | Endothermic | Surroundings cool |
| Photosynthesis | Endothermic | Requires light energy input |
Feel the surroundings, not the system
A quick check: if a reaction feels hot to your hand, the system released heat *into* your hand — that's exothermic. If it feels cold, the system pulled heat *out of* your hand — that's endothermic. Your hand is part of the surroundings.
Enthalpy (ΔH) and sign conventions
Enthalpy (H) is the heat content of a system at constant pressure — most reactions run in open beakers or flasks exposed to the atmosphere, so constant pressure is the normal lab condition. You can't measure H directly, but you can measure the change in enthalpy, ΔH, which equals the heat absorbed or released by a reaction at constant pressure: ΔH = H(products) − H(reactants) ΔH is written for the reaction as it's balanced, in kJ or kJ/mol, and its sign tells you the direction of heat flow for the system.
| Quantity | Positive (+) | Negative (−) |
|---|---|---|
| q (heat transferred to the system) | System gains heat → endothermic | System loses heat → exothermic |
| ΔH (enthalpy change) | H(products) > H(reactants) → endothermic | H(products) < H(reactants) → exothermic |
| ΔT (Tfinal − Tinitial) | Temperature increased | Temperature decreased |
The sign belongs to the system
ΔH and q are always reported from the system's perspective, never the surroundings'. A negative ΔH means the system lost energy (exothermic), even though the surroundings *gained* heat and got warmer. Mixing up whose perspective the sign describes is the single most common thermochemistry error.
At constant pressure, the heat measured in a calorimetry experiment IS the enthalpy change: q = ΔH. That's why q and ΔH share the same sign convention above, and why the next section's q = mcΔT equation is really how ΔH gets measured in the lab.
Specific heat and q = mcΔT
Specific heat capacity (c), usually just called specific heat, is the amount of heat needed to raise the temperature of 1 gram of a substance by 1 °C. It's a property of the material — metals have low specific heat (they heat up and cool down fast), water has unusually high specific heat (it resists temperature change), which is why water is used to regulate temperature in everything from car radiators to your own body. The governing equation connects heat, mass, specific heat, and temperature change:
q = m c ΔT
q = heat transferred (J or kJ) · m = mass (g) · c = specific heat [J/(g·°C)] · ΔT = Tfinal − Tinitial (°C). Always subtract *final minus initial* — reversing the order flips the sign and turns a correct exothermic answer into an endothermic one, or vice versa.
Worked example — heating water. How much heat is needed to raise 50.0 g of water from 20.0 °C to 80.0 °C? q = m c ΔT = (50.0 g)(4.184 J/g·°C)(80.0 °C − 20.0 °C) q = (50.0)(4.184)(60.0) = 12 552 J ≈ 12.6 kJ ΔT is positive (temperature rose), so q is positive — the water absorbed heat, consistent with something being heated.
Specific heat reference table + calorimetry
Specific heat values for common substances, in J/(g·°C):
| Substance | Specific heat, J/(g·°C) |
|---|---|
| Water (liquid) | 4.184 |
| Water (ice) | 2.09 |
| Water (steam) | 2.03 |
| Ethanol | 2.44 |
| Aluminum | 0.897 |
| Glass | 0.84 |
| Iron | 0.449 |
| Copper | 0.385 |
| Gold | 0.129 |
Calorimetry problems put two substances at different temperatures in contact and use conservation of energy: heat lost by the hotter substance equals heat gained by the cooler one, so q(lost) = −q(gained). Worked example. A 20.0 g piece of aluminum at 100.0 °C is dropped into 100.0 g of water at 20.0 °C. Assuming no heat escapes to the surroundings, what is the final temperature? Set q(Al) = −q(water): m(Al) c(Al) (Tf − 100.0) = −m(water) c(water) (Tf − 20.0) (20.0)(0.897)(Tf − 100.0) = −(100.0)(4.184)(Tf − 20.0) 17.94(Tf − 100.0) = −418.4(Tf − 20.0) 17.94Tf − 1794 = −418.4Tf + 8368 436.34Tf = 10 162 Tf ≈ 23.3 °C Notice the final temperature lands much closer to the water's starting temperature than the aluminum's — water's high specific heat and larger mass both dominate the outcome.
Match your units
If a specific heat is given in J/(g·°C), mass must be in grams, not kilograms. Mixing g and kg (or forgetting to convert) is a very common unit error and throws the answer off by a factor of 1000.
Heating curves
A heating curve plots temperature (y-axis) against heat added (x-axis) as a substance is heated steadily from solid through liquid to gas. It has five segments that alternate between sloped and flat:
| Segment | What's happening | Temperature behavior |
|---|---|---|
| 1. Solid warming | Kinetic energy of solid particles increases | Rises |
| 2. Melting (solid ↔ liquid) | Energy breaks the solid's fixed structure | Flat plateau at the melting point |
| 3. Liquid warming | Kinetic energy of liquid particles increases | Rises |
| 4. Boiling (liquid ↔ gas) | Energy fully separates particles into gas | Flat plateau at the boiling point |
| 5. Gas warming | Kinetic energy of gas particles increases | Rises |
During the two flat plateaus, energy is still being added, but it goes into breaking intermolecular forces (overcoming attractions between particles) rather than increasing kinetic energy — so temperature holds steady even though heating continues. The energy needed for these phase changes is called latent heat: the heat of fusion (ΔHfus) for melting/freezing, and the heat of vaporization (ΔHvap) for boiling/condensing. For water: ΔHfus = 334 J/g (6.01 kJ/mol) at 0 °C, and ΔHvap = 2260 J/g (40.7 kJ/mol) at 100 °C. Vaporization always takes far more energy than fusion for the same substance, because boiling completely separates particles while melting only loosens their fixed arrangement.
Which equation applies where
On a sloped segment (warming within one phase), use q = mcΔT with that phase's specific heat. On a flat segment (a phase change), use q = m·ΔHfus or q = m·ΔHvap instead — ΔT is zero there, so q = mcΔT would incorrectly give zero heat for a step that clearly takes real energy.
Try it yourself
calorimetry-trials.csvDownload calorimetry-trials.csv: https://restedteacher.com/practice-files/chemistry/calorimetry-trials.csv
Six calorimetry trials, each a substance with a mass, a specific heat, a starting temperature, and an ending temperature.
Open it as a spreadsheet
Open calorimetry-trials.csv in Google Sheets or Excel. Add a ΔT column and a q column (q = mass x specific heat x ΔT) and fill both down for all six trials at once.
Using the values in the file (q = mcΔT, with ΔT = final temperature minus initial temperature), answer these. Round q to 3 significant figures. 1. Calculate q for trial 6. Is the process exothermic or endothermic for the substance? 2. Which trial released the greatest amount of heat? 3. Trials 1, 3, and 5 are all endothermic. Rank those three trials from most heat absorbed to least.