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Calorimetry & Hess's Law — Advanced Problem-Solving

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When two substances share heat, energy is conserved

The introductory calorimetry problems heated one substance. Exam problems usually put two substances in thermal contact — a hot metal dropped into cool water — inside an insulated cup. No heat escapes, so the heat lost by the hotter object equals the heat gained by the cooler one: q(metal) + q(water) = 0, or equivalently −q(metal) = q(water). Both objects finish at the same final temperature T(f), and each q is computed from its own m·c·ΔT with ΔT = T(f) − T(initial).

Heat and temperature change
q = m·c·ΔT with q(hot) + q(cold) = 0 (insulated)
q = heat (J), m = mass (g), c = specific heat (J·g⁻¹·°C⁻¹), ΔT = T(final) − T(initial). The hot object has ΔT < 0 (q < 0); the cold object has ΔT > 0 (q > 0).
Worked example

A 55.0 g piece of copper at 99.0 °C is dropped into 100.0 g of water at 22.0 °C in an insulated calorimeter. Find the final temperature. c(Cu) = 0.385 J·g⁻¹·°C⁻¹, c(H₂O) = 4.18 J·g⁻¹·°C⁻¹.

  1. 1.Conservation: heat lost by copper = heat gained by water, so m(Cu)·c(Cu)·(99.0 − T(f)) = m(H₂O)·c(H₂O)·(T(f) − 22.0).
  2. 2.Evaluate the heat capacities: m(Cu)·c(Cu) = 55.0 × 0.385 = 21.175 J·°C⁻¹; m(H₂O)·c(H₂O) = 100.0 × 4.18 = 418 J·°C⁻¹.
  3. 3.Substitute: 21.175(99.0 − T(f)) = 418(T(f) − 22.0), i.e. 2096.3 − 21.175 T(f) = 418 T(f) − 9196.
  4. 4.Collect T(f): 2096.3 + 9196 = (418 + 21.175) T(f), so 11292.3 = 439.175 T(f).
  5. 5.Divide: T(f) = 11292.3 ÷ 439.175 = 25.7 °C.
Answer: T(f) ≈ 25.7 °C — close to the water's starting temperature, because water's large mass and specific heat dominate the balance.

Bomb calorimeters measure heat with a single heat capacity

A bomb calorimeter burns a sample in a sealed, rigid steel vessel (constant volume) submerged in water. Rather than tracking the water and hardware separately, the whole assembly is characterized by one calorimeter heat capacity C (units kJ·°C⁻¹), found beforehand by burning a standard. The heat the calorimeter absorbs is simply q(cal) = C·ΔT. Because the burn releases that heat, q(reaction) = −q(cal). Dividing by the moles of sample converts it to a molar ΔH.

Worked example

Burning 1.00 g of glucose (M = 180.2 g·mol⁻¹) in a bomb calorimeter of heat capacity C = 6.50 kJ·°C⁻¹ raises the temperature from 23.00 °C to 25.40 °C. Find the molar enthalpy of combustion.

  1. 1.Temperature change: ΔT = 25.40 − 23.00 = 2.40 °C.
  2. 2.Heat absorbed by the calorimeter: q(cal) = C·ΔT = 6.50 kJ·°C⁻¹ × 2.40 °C = 15.6 kJ.
  3. 3.The reaction released this heat, so q(reaction) = −15.6 kJ for the 1.00 g burned.
  4. 4.Convert 1.00 g to moles: 1.00 g ÷ 180.2 g·mol⁻¹ = 0.005549 mol.
  5. 5.Divide: ΔH = −15.6 kJ ÷ 0.005549 mol = −2811 kJ·mol⁻¹.
Answer: ΔH ≈ −2.81 × 10³ kJ·mol⁻¹ (exothermic), matching the accepted glucose value of about −2803 kJ·mol⁻¹.

Hess's law: assemble a target from known reactions

Because enthalpy is a state function, ΔH for a target reaction equals the sum of the ΔH values of any set of steps that add up to it. Two moves do all the work: reverse a reaction (flip the sign of its ΔH) and scale a reaction (multiply the whole equation and its ΔH by a factor). Choose the moves so that every species not in the target cancels, then add the adjusted ΔH values.

Two cross-checks for ΔH°rxn
ΔH°rxn = Σ ΔH°f(products) − Σ ΔH°f(reactants); ΔH ≈ Σ(bonds broken) − Σ(bonds formed)
Each ΔH°f is weighted by its coefficient; ΔH°f of an element in its standard state is 0. The bond-enthalpy estimate is gas-phase only and approximate (tabulated bonds are averages).
Worked example

Find ΔH for 2 C(s) + H₂(g) → C₂H₂(g) using: (1) C(s) + O₂(g) → CO₂(g), ΔH₁ = −393.5 kJ; (2) H₂(g) + ½ O₂(g) → H₂O(l), ΔH₂ = −285.8 kJ; (3) 2 C₂H₂(g) + 5 O₂(g) → 4 CO₂(g) + 2 H₂O(l), ΔH₃ = −2599 kJ.

  1. 1.The target needs 2 C(s), so scale reaction (1) by 2: 2 C + 2 O₂ → 2 CO₂, ΔH = 2(−393.5) = −787.0 kJ.
  2. 2.The target needs 1 H₂, so keep reaction (2) as written: H₂ + ½ O₂ → H₂O, ΔH = −285.8 kJ.
  3. 3.The target makes 1 C₂H₂ as a product, but (3) consumes 2 C₂H₂, so reverse and halve (3): 2 CO₂ + H₂O → C₂H₂ + 5/2 O₂, ΔH = −½(−2599) = +1299.5 kJ.
  4. 4.Add the three: the 2 CO₂ and the H₂O cancel, and the O₂ terms cancel (2 + ½ on the left match 5/2 on the right), leaving 2 C + H₂ → C₂H₂.
  5. 5.Sum the ΔH values: −787.0 + (−285.8) + 1299.5 = +226.7 kJ.
Answer: ΔH = +226.7 kJ (endothermic) — acetylene is an energy-rich, unstable-to-form molecule, consistent with the accepted +227 kJ·mol⁻¹.
Watch out

Reversing a reaction flips the sign of ΔH; scaling by a factor multiplies ΔH by that same factor. If you do both to one equation, do both to its ΔH: reverse-and-halve reaction (3) means ΔH becomes −(½)(ΔH₃), not just half of it.

Checkpoint

A sample is burned in a bomb calorimeter of heat capacity C = 4.90 kJ·°C⁻¹, and the temperature rises from 25.00 °C to 29.00 °C. How much heat did the reaction release?

Checkpoint

Given N₂(g) + 3 H₂(g) → 2 NH₃(g), ΔH = −92 kJ, what is ΔH for NH₃(g) → ½ N₂(g) + 3/2 H₂(g)?

Checkpoint

Estimate ΔH for CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(g) from bond enthalpies (kJ·mol⁻¹): C−H = 414, O=O = 498, C=O = 799, O−H = 463.

On the exam

On the free-response section, calorimetry usually feeds Hess or ΔH°f: measure q(reaction) with the calorimeter, divide by moles to get a molar ΔH, then use it as one step or cross-check it against Σ ΔH°f(products) − Σ ΔH°f(reactants). Always report the final sign — exothermic is negative.

Answer the 3 checkpoints as you read.

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