6.4 Heat Capacity and Calorimetry

Heat capacity (C) is the amount of heat required to raise the temperature of an object or sample by 1°C (units: J/°C or J/K). Specific heat capacity (c, often just “specific heat”) is an intensive property: the heat required to raise 1 gram of a substance by 1°C (J/g·°C). Use q = m c ΔT when you know mass and specific heat—that’s the AP-tested formula for calculating heat transferred (CED 6.4.A.1). If you instead know the whole object’s heat capacity, use q = C ΔT. Related: molar heat capacity is heat per mole per degree (J/mol·°C). In calorimetry, you often measure temperature change and use specific (or molar) heats, plus calorimeter constants for heat absorbed by the apparatus. For a focused review, see the Topic 6.4 study guide (https://library.fiveable.me/ap-chemistry/unit-6/heat-capacity-calorimetry/study-guide/jShImkrhZMnPWxlEjdwN). More practice problems across Unit 6 are at (https://library.fiveable.me/practice/ap-chemistry).

Use q = mcΔT to calculate the heat gained or lost by a substance: q (J) = m (g) × c (J/g·°C) × ΔT (°C, = Tfinal − Tinitial). Steps you’ll use on AP problems (Topic 6.4): 1. Identify which body you’re finding q for (usually water or the system). Use mass in grams and the correct specific heat (water ≈ 4.184 J/g·°C unless told otherwise). 2. Compute ΔT = Tfinal − Tinitial. If T rises, ΔT positive → q positive (system gained heat). If T falls, ΔT negative → q negative (system lost heat). 3. Plug into q = mcΔT and keep units (J). For calorimetry, apply conservation of energy: qlost + qgained = 0 (or qsystem = −qsurroundings). 4. If asked per mole, convert mass → moles and use molar heat capacity or divide q by moles. On the exam you may need the calorimeter constant or include calorimeter heat (qcal = CcalΔT). For review and worked examples see the Topic 6.4 study guide (https://library.fiveable.me/ap-chemistry/unit-6/heat-capacity-calorimetry/study-guide/jShImkrhZMnPWxlEjdwN) and extra practice (https://library.fiveable.me/practice/ap-chemistry).

Water’s specific heat is high (about 4.18 J/g·°C) because a lot of the added thermal energy goes into changing intermolecular interactions rather than immediately raising kinetic energy (temperature). Water is small and highly polar, so molecules form a dense network of hydrogen bonds. To raise water’s temperature you must partially overcome and excite those H-bonded arrangements (stretching/rotational/vibrational modes), which consumes energy per gram more than in substances with weaker intermolecular forces. That’s why, for the same q in q = mcΔT, ΔT is smaller for water (CED 6.4.A.1 & 6.4.A.3). Practically: water moderates temperature changes in calorimetry and is why it’s commonly the calorimeter medium. For more review on specific heat and calorimetry examples, see the Topic 6.4 study guide (https://library.fiveable.me/ap-chemistry/unit-6/heat-capacity-calorimetry/study-guide/jShImkrhZMnPWxlEjdwN) and the Unit 6 overview (https://library.fiveable.me/ap-chemistry/unit-6). Want extra practice? Try the AP Chem problem set list (https://library.fiveable.me/practice/ap-chemistry).

Coffee-cup calorimetry is just a simple way to measure heat transfer at constant pressure (like in many AP problems). You mix a hot (or reactive) sample into water inside an insulated cup, wait until thermal equilibrium, measure the temperature change ΔT, and use q = mcΔT to find the heat gained or lost. Because of energy conservation (First Law), q(system) + q(surroundings) = 0—so if the solution/water warms by qwater = mwater·cwater·ΔT, the reaction or added object released qreaction = −qwater. Remember to include heat absorbed by the calorimeter itself if it’s not ideal (use a calorimeter constant or include the mass & specific heat of the cup/spoon). Use cwater ≈ 4.18 J/g·°C unless told otherwise. On the AP exam you’ll be asked to calculate sign (exothermic vs endothermic), use specific/molar heat as needed, and show energy bookkeeping. For a focused review, see the Topic 6.4 study guide (https://library.fiveable.me/ap-chemistry/unit-6/heat-capacity-calorimetry/study-guide/jShImkrhZMnPWxlEjdwN). Want more practice? Try the AP Chem practice set (https://library.fiveable.me/practice/ap-chemistry).

Specific heat capacity (usually c) is the energy required to raise 1 gram of a substance by 1°C: units J/(g·°C). Molar heat capacity (Cm) is the energy required to raise 1 mole of a substance by 1°C: units J/(mol·°C). Both show up in AP Thermochemistry problems (CED 6.4.A)—use whichever matches the amount you have: - If you know mass: q = m c ΔT (q in J). - If you know moles: q = n Cm ΔT. They’re directly related: Cm = M × c, where M is molar mass (g/mol). Example: water c ≈ 4.18 J/(g·°C); M = 18.02 g/mol → Cm ≈ 18.02×4.18 ≈ 75.3 J/(mol·°C). On the AP exam pick the form that fits the given quantity and units (you’ll see both forms in questions). For a focused refresher, see the Topic 6.4 study guide (https://library.fiveable.me/ap-chemistry/unit-6/heat-capacity-calorimetry/study-guide/jShImkrhZMnPWxlEjdwN) and more practice problems (https://library.fiveable.me/practice/ap-chemistry).

Look at the temperature change of the calorimeter contents and use q = mcΔT (plus any calorimeter constant). Simple rules: - If the measured temperature of the solution/mixture rises, the process released heat → reaction is exothermic. - If the temperature falls, the process absorbed heat → reaction is endothermic. Do the math: calculate q_surroundings = m·c·ΔT (include the calorimeter’s heat capacity if given). By energy conservation (1st law), q_system + q_surroundings = 0, so q_reaction = −q_surroundings. Example: 100.0 g water (c = 4.18 J/g·°C) with ΔT = +5.00 °C gives q_water = 100·4.18·5 = 2090 J, so q_reaction = −2090 J (exothermic). For constant-pressure coffee-cup calorimetry, ΔH ≈ q_p. This is exactly the type of reasoning AP asks you to use (CED 6.4.A; q = mcΔT). For a quick topic review see the Topic 6.4 study guide (https://library.fiveable.me/ap-chemistry/unit-6/heat-capacity-calorimetry/study-guide/jShImkrhZMnPWxlEjdwN) and try practice problems (https://library.fiveable.me/practice/ap-chemistry).

The first law of thermodynamics—energy is conserved—is the foundation for heat-capacity and calorimetry calculations. In practice that means the heat lost by one part of the experiment equals the heat gained by the other: q(system) + q(surroundings) = 0. So when you use q = m c ΔT (CED 6.4.A.1) to find the heat change of a substance, you’re applying the first law to track energy flow. In a coffee-cup calorimeter you usually set q(reaction) = −q(water + calorimeter); in a bomb calorimeter you account for the calorimeter constant (q = Ccal ΔT). Watch sign convention: ΔT positive → q absorbed (endothermic), ΔT negative → q released (exothermic) (CED 6.4.A.2, 6.4.A.7). For AP: be ready to calculate q with specific or molar heat capacity and justify energy conservation in your answer (Topic 6.4). Review worked examples in the Topic 6.4 study guide (https://library.fiveable.me/ap-chemistry/unit-6/heat-capacity-calorimetry/study-guide/jShImkrhZMnPWxlEjdwN) and practice problems (https://library.fiveable.me/practice/ap-chemistry).

Because different materials store different amounts of thermal energy per gram, equal masses heat at different rates. The heat equation q = mcΔT tells you why: for the same heat input q and the same mass m, the temperature change ΔT = q/(m c). A large specific heat c means the substance needs more energy to raise its temperature, so ΔT is smaller (it heats more slowly). Example: water’s c ≈ 4.18 J/g·°C, many metals have c ≈ 0.1–1 J/g·°C—the metal will warm much faster than the same mass of water for the same heat added. On a molecular level, specific heat depends on how energy is stored (translational, rotational, vibrational modes), bond strength, and structure. In calorimetry you use q = mcΔT to measure heat flows and find c or ΔH; practice these setups for the AP exam (topic 6.4). For a clear study guide on this topic, see Fiveable’s heat-capacity/calorimetry page (https://library.fiveable.me/ap-chemistry/unit-6/heat-capacity-calorimetry/study-guide/jShImkrhZMnPWxlEjdwN). For more practice questions, check (https://library.fiveable.me/practice/ap-chemistry).

Use q = mcΔT. Here q is heat (J), m is mass (g), c is specific heat (J/g·°C) or use molar heat capacity (J/mol·°C) with moles, and ΔT = Tfinal − Tinitial (°C or K). Positive q = heat absorbed (endothermic); negative q = heat released (exothermic). In calorimetry problems remember energy is conserved: qsystem + qsurroundings = 0, so qsystem = −qsurroundings. Example: heat needed to warm 50.0 g of water (c = 4.18 J/g·°C) from 20.0 °C to 80.0 °C: q = (50.0 g)(4.18 J/g·°C)(80.0−20.0 °C) = 12,540 J (absorbed). On the AP exam you'll be asked to calculate q from mass or moles and c (CED 6.4.A, use q = mcΔT). For calorimetry labs, use q lost by hot object = −q gained by cold object and include the calorimeter constant if given. For more examples and aligned study notes, see the Topic 6.4 study guide (https://library.fiveable.me/ap-chemistry/unit-6/heat-capacity-calorimetry/study-guide/jShImkrhZMnPWxlEjdwN) and practice problems (https://library.fiveable.me/practice/ap-chemistry).

Short answer: usually, but not always. q = mcΔT is the go-to for heating/cooling a single substance (mass m, specific heat c, temp change ΔT)—that’s Essential Knowledge 6.4.A.1. But real calorimetry problems often need extra pieces: - Add/subtract heat terms: qreaction + qwater + qcalorimeter = 0. For a coffee-cup calorimeter you’ll use qwater = mcΔT and qcalorimeter = Ccal·ΔT (calorimeter constant), then set qrxn = −(qwater + qcalorimeter). - Phase changes use q = m·ΔHfus or m·ΔHvap (latent heat) not mcΔT. - Use molar heat capacity (Cmol) when given moles: q = n·Cmol·ΔT. - Bomb calorimeter: heat measured at constant volume; use qrxn = −(Cbomb·ΔT) or include calorimeter heat capacity plus water. On the AP exam, show conservation of energy (first law) and label signs (+/-) clearly. For a focused study guide and practice problems, see the Topic 6.4 study guide (https://library.fiveable.me/ap-chemistry/unit-6/heat-capacity-calorimetry/study-guide/jShImkrhZMnPWxlEjdwN) and Unit 6 overview (https://library.fiveable.me/ap-chemistry/unit-6). For lots of practice, try Fiveable’s AP Chem practice set (https://library.fiveable.me/practice/ap-chemistry).

When you heat a system, thermal energy is transferred into it—the system’s internal energy increases and its temperature usually rises; when you cool a system, thermal energy leaves it—its internal energy decreases and temperature falls. Quantitatively AP expects you to use q = mcΔT (q positive when the system absorbs heat, negative when it releases heat) and remember the first law: energy is conserved (ΔE = q + w). In calorimetry you measure these transfers (coffee-cup for constant-pressure, bomb calorimeter for constant-volume) and use temperature changes plus specific or molar heat capacities to calculate q (Topic 6.4 / LO 6.4.A). If a dissolution raises the calorimeter temperature the process is exothermic (releases heat); if it lowers temperature it’s endothermic (absorbs heat). For practice, review the Topic 6.4 study guide (https://library.fiveable.me/ap-chemistry/unit-6/heat-capacity-calorimetry/study-guide/jShImkrhZMnPWxlEjdwN) and try problems at (https://library.fiveable.me/practice/ap-chemistry).

Phase changes break the simple q = mcΔT rule because temperature doesn't change while a phase transition happens. For heating/cooling you use q = mcΔT (CED 6.4.A.1, 6.4.A.5), but for melting/boiling you must use latent (transition) heat: q = m·L (L = heat of fusion or vaporization). In calorimetry problems treat the process in steps—heat or cool to the phase-change temperature (use q = mcΔT), then apply q = mL for the phase change (ΔT = 0), then continue heating/cooling the new phase with its specific heat. Sum all q terms (and include calorimeter heat if given) and conserve energy (first law) to solve. Remember specific vs. molar heat capacities and include correct units. For guided examples and AP-style practice, see the Topic 6.4 study guide (https://library.fiveable.me/ap-chemistry/unit-6/heat-capacity-calorimetry/study-guide/jShImkrhZMnPWxlEjdwN) and more practice problems (https://library.fiveable.me/practice/ap-chemistry).

When you dissolve salt in water the temperature changes because energy is transferred between the solution and its surroundings as bonds are broken and formed—that’s the enthalpy of solution (ΔHsoln). Breaking the ionic lattice (and hydrating ions) requires energy; forming ion–water interactions releases energy. If the net process releases more energy than it consumes, the solution heats up (exothermic); if it absorbs more, the solution cools (endothermic). In a calorimeter the heat lost or gained by the solution equals q = mcΔT (CED 6.4.A.1, 6.4.A.7), so the measured ΔT tells you the direction and magnitude of heat flow. Remember energy is conserved (first law), so the temperature change reflects the system reaching thermal equilibrium after that energy exchange. For a quick review, see the Topic 6.4 study guide (https://library.fiveable.me/ap-chemistry/unit-6/heat-capacity-calorimetry/study-guide/jShImkrhZMnPWxlEjdwN) and more practice problems (https://library.fiveable.me/practice/ap-chemistry).

There are three main ways a chemical system changes its energy (CED 6.4.A.6): - Heating/cooling (thermal energy transfer): energy flows into or out of a system as heat, changing temperature. You calculate this with q = mcΔT using specific or molar heat capacity (CED 6.4.A.1, 6.4.A.5). Calorimetry measures these transfers. - Phase transitions (latent heat): energy is absorbed or released without a temperature change when a substance melts, vaporizes, etc. Use latent heats (ΔHfusion, ΔHvap) in energy bookkeeping (related Topic 6.5). - Chemical reactions (bond making/breaking): reaction enthalpies (ΔHrxn) change system energy—exothermic reactions release heat, endothermic absorb it (CED 6.4.A.7, 6.4.A.2). On the AP exam you’ll be asked to calculate q for heating/cooling problems, infer direction of heat flow in calorimetry, and connect phase-change or reaction enthalpies to temperature changes (see the Topic 6.4 study guide for worked examples: https://library.fiveable.me/ap-chemistry/unit-6/heat-capacity-calorimetry/study-guide/jShImkrhZMnPWxlEjdwN). For broader review, check the Unit 6 overview (https://library.fiveable.me/ap-chemistry/unit-6) and drill practice problems (https://library.fiveable.me/practice/ap-chemistry).

Look at the temperature change of the solution and use q = mcΔT plus energy conservation. Steps: measure Tinitial and Tfinal of the calorimeter mixture, find ΔT = Tfinal − Tinitial, then calculate q(solution) = msolution · c · ΔT (use cwater ≈ 4.18 J/g·°C if water is the solvent). If the calorimeter has a known heat capacity, add q(cal) = Ccal · ΔT and include that in the surroundings’ heat. By the first law, qsystem = −qsurroundings. So: - If Tfinal > Tinitial → ΔT positive → q(solution) positive → solution gained heat → dissolution released heat (exothermic; qsystem < 0). - If Tfinal < Tinitial → ΔT negative → q(solution) negative → solution lost heat → dissolution absorbed heat (endothermic; qsystem > 0). This aligns with CED 6.4.A (q = mcΔT, thermal energy flow, conservation of energy). For practice, see the Topic 6.4 study guide (https://library.fiveable.me/ap-chemistry/unit-6/heat-capacity-calorimetry/study-guide/jShImkrhZMnPWxlEjdwN) and more problems (https://library.fiveable.me/practice/ap-chemistry).