Heat Capacity and Calorimetry: Study Guide (Chemistry 6.4)

From direction to amount

In Topic 6.3 you explained which way energy moves when a warmer body meets a cooler one. This topic asks how much.

The amount of energy transferred as heat has the symbol q and is measured in joules (J) or kilojoules (1 kJ = 1000 J). Its sign tells you the direction for the system you have chosen:

• q positive: the system absorbs heat. Heating a system increases its energy.
• q negative: the system releases heat. Cooling a system decreases its energy.

A system can change its energy in three main ways: by heating or cooling (this topic), by a phase change (Topic 6.5) and by a chemical reaction (Topics 6.6 to 6.9). Here the substance stays in the same phase and nothing reacts; only its temperature changes.

Specific heat capacity and q = mcΔT

How much energy does it take to warm something up? It depends on three things:

1. How much substance there is (its mass, m). Twice the mass needs twice the energy for the same temperature rise.
2. How big the temperature change is (ΔT). A 20 °C rise needs twice the energy of a 10 °C rise.
3. What the substance is. This is captured by its specific heat capacity, c: the energy needed to raise the temperature of 1 g of the substance by 1 °C.

Putting these together gives the heat transfer equation:

q = mcΔT, where q is in J, m in g, c in J g⁻¹ °C⁻¹ and ΔT = T(final) − T(initial) in °C.

Because 1 °C and 1 K are the same size, the units J g⁻¹ °C⁻¹ and J g⁻¹ K⁻¹ mean exactly the same thing. ΔT carries the sign: a temperature rise gives a positive q (heat absorbed), a fall gives a negative q (heat released).

Substance (at about 25 °C)	c (J g⁻¹ °C⁻¹)
liquid water	4.18
ethanol	2.44
aluminium	0.897
iron	0.449
copper	0.385

Water’s specific heat capacity is unusually large. That is why it is used to carry energy in heating systems, and why the water in a calorimeter is a good “energy store” to measure with.

Equal masses, different temperature changes

Give the same amount of energy to equal masses of two substances and they will not reach the same temperature. Rearranging q = mcΔT:

ΔT = q / (mc)

For a fixed q and m, ΔT is inversely proportional to c. Figure 1 shows 100.0 g of water, aluminium and copper each receiving up to 2.00 kJ.

The steeper the line, the smaller the specific heat capacity. Two kilojoules warm 100.0 g of copper by about 52 °C but the same mass of water by only 4.78 °C.

Molar heat capacity and the heat capacity of an object

Specific heat capacity is “per gram”. Sometimes “per mole” is more useful.

• Molar heat capacity, C_m, is the energy needed to raise the temperature of 1 mol of a substance by 1 °C. Its unit is J mol⁻¹ °C⁻¹. To convert, multiply by the molar mass: C_m = c × M. Then q = n C_m ΔT.
• Heat capacity of a whole object, C, is the energy needed to raise the temperature of that object by 1 °C. Its unit is J °C⁻¹. For an object made of one substance, C = mc, so q = CΔT. A 250 g aluminium pan has C = 250 × 0.897 = 224 J °C⁻¹.

Look what happens when you convert the metals in the table to molar heat capacities:

Metal	c (J g⁻¹ °C⁻¹)	M (g mol⁻¹)	C_m = c × M (J mol⁻¹ °C⁻¹)
aluminium	0.897	26.98	24.2
iron	0.449	55.85	25.1
copper	0.385	63.55	24.5

Per gram the values differ by more than a factor of two, but per mole they are all close to 25 J mol⁻¹ °C⁻¹. A gram of aluminium contains more than twice as many atoms as a gram of copper, so it needs more energy for the same temperature rise. This is a useful particle-level link: for these simple metals, the energy needed per atom is similar.

Calorimetry: energy lost = energy gained

The first law of thermodynamics states that energy is conserved in every chemical and physical process. It is not created or destroyed, only transferred. Calorimetry uses this law to measure heat transfer.

A simple coffee-cup calorimeter (Figure 2) is two nested polystyrene cups with a lid, a thermometer and a stirrer. Polystyrene is a poor conductor, so very little energy enters or leaves; the lid cuts losses to the air.

Inside an insulated calorimeter, all the energy released by one part is absorbed by the other parts:

q(released) + q(absorbed) = 0, or q(lost) = −q(gained)

For a hot object dropped into cool water, q(object) + q(water) = 0. For a process such as dissolving, q(process) + q(solution) = 0, so q(process) = −q(solution).

Direction of energy flow when something dissolves

When a solid dissolves in water inside a calorimeter, the thermometer reads the temperature of the mixture:

• Temperature rises: the solution gained energy, so the dissolving process released energy. Dissolving is exothermic, and q(dissolving) is negative.
• Temperature falls: the solution lost energy, so the dissolving process absorbed energy. Dissolving is endothermic, and q(dissolving) is positive.

The usual assumptions

Calorimetry calculations rely on stated assumptions. Name them when a question asks about sources of error.

1. No energy is exchanged with the outside (perfect insulation).
2. The calorimeter itself absorbs negligible energy, unless you are given its heat capacity.
3. A dilute solution has the specific heat capacity of water, 4.18 J g⁻¹ °C⁻¹, and the mass of the solution is the mass of the water plus the mass of the solute.
4. The final temperature is the equilibrium temperature, reached after stirring (Topic 6.3).

Worked example 1: heating water, and the same energy in a metal

Question. (a) How much energy is needed to heat 275 g of water from 19.0 °C to 85.0 °C? (b) If the same energy were given to 275 g of aluminium starting at 19.0 °C, what would its temperature rise be?

1. ΔT for the water: 85.0 − 19.0 = 66.0 °C.
2. q = mcΔT = 275 g × 4.18 J g⁻¹ °C⁻¹ × 66.0 °C = 75 867 J.
3. Answer (a): q = 75.9 kJ (positive: the water absorbs energy).
4. (b) Rearrange: ΔT = q / (mc) = 75 867 J ÷ (275 g × 0.897 J g⁻¹ °C⁻¹) = 308 °C (307.6 °C).

Interpretation. Water’s c is 4.18 / 0.897 = 4.66 times aluminium’s, so the same energy warms the same mass of aluminium about 4.66 times as much. Equal masses with different specific heat capacities do not change temperature by the same amount.

Worked example 2: finding the specific heat capacity of a metal

Question. A 62.0 g sample of an unknown metal Z is heated to 99.5 °C in boiling water, then quickly moved into an insulated calorimeter containing 75.0 g of water at 21.0 °C. After stirring, the final temperature is 26.6 °C. Find the specific heat capacity of Z. Assume the calorimeter absorbs negligible energy.

1. Water: ΔT = 26.6 − 21.0 = 5.6 °C. q(water) = 75.0 × 4.18 × 5.6 = +1755.6 J (absorbed).
2. Conservation of energy: q(metal) + q(water) = 0, so q(metal) = −1755.6 J (released).
3. Metal: ΔT = 26.6 − 99.5 = −72.9 °C.
4. Solve for c: c = q / (mΔT) = −1755.6 J ÷ (62.0 g × −72.9 °C) = 0.3884 J g⁻¹ °C⁻¹.

Answer. c(Z) = 0.39 J g⁻¹ °C⁻¹. Of the metals in the table, this matches copper (0.385) and rules out iron (0.449) and aluminium (0.897).

Attending to precision. The water’s ΔT, 5.6 °C, has only two significant figures, even though each thermometer reading has three. Subtracting two readings loses precision. So the answer is given to 2 significant figures. A larger mass of metal or a smaller mass of water would give a bigger ΔT and a more precise result.

Worked example 3: a dissolution calorimeter

Question. 5.10 g of a fictional salt QZ (M = 85.0 g mol⁻¹) is dissolved in 50.0 g of water in a coffee-cup calorimeter. The temperature falls from 22.4 °C to 16.9 °C. (a) Is dissolving QZ endothermic or exothermic? (b) Calculate q for the dissolving process. (c) Calculate the energy absorbed per mole of QZ.

1. (a) Direction. The temperature of the mixture fell, so energy moved from the solution into the dissolving process. Dissolving QZ is endothermic.
2. Mass of solution: 50.0 + 5.10 = 55.10 g. ΔT = 16.9 − 22.4 = −5.5 °C.
3. q(solution) = 55.10 × 4.18 × (−5.5) = −1266.7 J (the solution lost energy).
4. (b) q(dissolving) = −q(solution) = +1.3 × 10³ J (1266.7 J). Positive, as expected for an endothermic process.
5. (c) Moles of QZ: n = 5.10 ÷ 85.0 = 0.0600 mol. Energy per mole = 1266.7 J ÷ 0.0600 mol = 21 112 J mol⁻¹ = +21 kJ mol⁻¹.

Assumptions. No energy exchanged with the air or the cup; the solution’s specific heat capacity is 4.18 J g⁻¹ °C⁻¹. If you used the mass of water only (50.0 g), you would get 19 kJ mol⁻¹; either is accepted if stated, but use the total mass unless told otherwise. In Topic 6.6 this energy per mole, measured at constant pressure, becomes the enthalpy of solution, ΔH.

Common misconceptions

• “A high specific heat capacity means it heats up quickly.” The opposite. A large c means more energy is needed for each degree, so the temperature rises less.
• “The temperature rose, so the dissolving absorbed heat.” The thermometer is in the solution (the surroundings). A rise means the process released energy into the solution.
• “q of the water is q of the reaction.” They are equal in size and opposite in sign: q(process) = −q(solution).
• “ΔT must be converted to kelvin.” A change of 1 °C equals a change of 1 K, so ΔT is the same number in both.
• “The final temperature is halfway between the starting temperatures.” Only if the two parts have equal heat capacities (mc). Otherwise it lies closer to the starting temperature of the part with the larger mc.
• “Heat and temperature are the same.” Temperature measures the average kinetic energy of the particles; q is energy transferred. A full bath at 40 °C releases far more heat as it cools to room temperature than a small cup of tea at 80 °C, because it has a much larger mass.
• Mixing up specific and molar heat capacity. Check the unit: per gram (J g⁻¹ °C⁻¹) goes with mass; per mole (J mol⁻¹ °C⁻¹) goes with moles.
• Keeping too many significant figures. The temperature change, found by subtraction, usually limits the precision.

Where this leads

Next, Topic 6.5 deals with the second way a system changes its energy: phase changes, where energy is absorbed or released with no change in temperature. The calorimetry method here is the basis for measuring enthalpies of reaction in Topic 6.6. Try the practice questions now, then use the revision notes and the checklist to consolidate.