Why this unit matters
Energy accounting links microscopic bond changes to measurable temperature change and process feasibility.
What you will learn
- Use q = mcDeltaT to calculate heat transfer in a simple calorimetry setup.
- Interpret endothermic and exothermic processes using enthalpy sign conventions.
- Apply Hess's law and standard enthalpies of formation to compute reaction enthalpy.
Understand the core ideas
Thermochemistry tracks energy transfer during physical and chemical change with clear system boundaries. Heat q is energy in transit caused by temperature difference, while temperature is a measure related to average particle kinetic energy. Sign conventions are essential: q is positive for a system that absorbs heat and negative for a system that releases heat. In coffee-cup calorimetry, a common AP assumption is that heat exchanged with the external environment is negligible, so heat lost by reaction equals heat gained by solution and cup combined. Most introductory calculations use q = mcDeltaT with mass in grams, specific heat in J g^-1 C^-1, and DeltaT in C. Keeping those units explicit prevents sign and scale mistakes. Conceptually, this chapter connects microscopic bond-energy changes to measurable temperature changes, which is why both arithmetic and interpretation are tested together. Define the system first to avoid sign errors in written explanations. Then label surroundings clearly before interpreting measured temperature changes.
Enthalpy change represents heat flow at constant pressure and allows reaction-level energy bookkeeping. Exothermic processes have negative , meaning the system releases heat, while endothermic processes have positive , meaning the system absorbs heat. Energy diagrams visualize these differences and help compare pathways with different activation barriers. Hess's law works because enthalpy is a state function, so total depends only on initial and final states, not on path. Practically, reversing an equation changes the sign of , and multiplying an equation by a factor multiplies by the same factor. Formation-enthalpy calculations follow the same logic through products minus reactants sums weighted by coefficients. Reliable AP solutions state assumptions, track signs carefully, and include a reasonableness check such as whether temperature change direction matches endothermic or exothermic interpretation for the defined system and surroundings. This makes multi-equation energy synthesis much more reliable. It also simplifies error tracing when one intermediate equation is reversed incorrectly.
Key terms
- specific heat capacity
- Heat required to raise 1 g of a substance by 1 degree C (or 1 K).
- enthalpy change
- Heat of reaction at constant pressure, written as .
- state function
- Property whose change depends only on initial and final states, not on the path taken.
- Hess's law
- Principle that reaction enthalpies add when chemical equations are added.
Compute heat absorbed by water in calorimetry
Assume 100.0 g water warms from 22.0 C to 27.0 C in an insulated cup. Use c = 4.184 J g^-1 C^-1. Treat water as the system for this calculation.
- 1) Compute temperature change: DeltaT = 27.0 - 22.0 = 5.0 C.
- 2) Apply q = mcDeltaT with units: q = (100.0 g)(4.184 J g^-1 C^-1)(5.0 C).
- 3) Multiply values: 100.0 x 4.184 x 5.0 = 2092 J.
- 4) Assign sign from temperature increase: q is positive for water because it gained heat.
A common misconception
Claim: If is negative, the reaction mixture temperature must always decrease.
Correction: Negative means the reacting system releases heat. If that heat is captured by surroundings such as solution in a cup, the measured surroundings temperature often increases.
Lessons in this unit
- System, surroundings, and energy flowDefine sign conventions and distinguish heat from temperature.
- Calorimetry calculationsSolve heat-transfer problems with mass, specific heat, and DeltaT.
- Enthalpy and reaction profilesUse energy diagrams to classify and compare reaction pathways.
- Hess's law and formation dataCombine equations or tabulated values to find for target reactions.
Study task
Unit checkpoint
A 100.0 g sample of water warms by 5.0 C. Using c = 4.184 J/(g C), what is q for the water?
q = mcDeltaT = (100.0 g)(4.184 J/(g C))(5.0 C) = 2092 J, positive for the water because it gains heat.