Heat Capacity and Temperature Change

Introduction

A thermometer reports temperature, not heat. Heat capacity links the measured temperature change of a sample or calorimeter to the heat it receives or loses. The relation q = CΔT is simple when C is approximately constant over the range, but signs, units and the identity of the warmed object must be clear.

Core explanation

The heat capacity of a specified object is the heat required to raise its temperature by one kelvin under a stated constraint and over a suitable small range. For a finite change with nearly constant C, q object = C(T f − T i) = CΔT. A warming object has ΔT > 0 and q > 0; a cooling object has ΔT < 0 and q < 0. The heat capacity C is positive for ordinary stable materials under familiar conditions.

Heat capacity is not the same as temperature. A large mass of water can absorb much more heat than a small mass while experiencing the same ΔT. The value C for a whole sample depends on its amount and material. One can also use specific heat capacity c per gram, giving q = mcΔT, or molar heat capacity C m per mole, giving q = nC mΔT. The next page compares these units in detail.

In calorimetry, the thermometer usually measures the surroundings of the reaction: solution, water bath or apparatus. If that medium warms, q medium is positive. For an approximately insulated combined system, the reaction loses corresponding heat, so q reaction ≈ −q medium after including all warmed parts. The thermometer rise is not itself the reaction enthalpy; heat capacity and reaction amount are needed.

The choice of constant-pressure or constant-volume heat capacity matters for gases because expansion work differs. A gas's C P and C V are generally not equal. For a liquid solution near atmospheric pressure, using a measured or approximated specific heat can estimate the heat absorbed by the solution. The chemical reaction energy is then related to ΔH if the process is suitably constant pressure and non-P–V work is negligible.

Heat capacity can change with temperature, phase and composition. The constant-C formula is an approximation over a stated interval. Heating through melting or boiling involves latent heat, so temperature can remain nearly constant while substantial heat is absorbed. A single q = CΔT term cannot cover the phase transition by itself.

Measurement quality matters. Heat can escape through imperfect insulation, stirrers and thermometers have their own heat capacities, and a delayed reading can miss the peak temperature. A calibrated calorimeter constant C cal accounts for apparatus heat absorption: q cal = C calΔT. A careful energy balance includes solution and apparatus terms separately if both are significant.

Unit consistency is essential. If C is in J K⁻¹ and ΔT in K, q is in J. A temperature difference of 5 °C equals a difference of 5 K, but an absolute temperature of 5 °C is not 5 K. Do not use degrees Celsius directly in formulas involving RT or absolute thermodynamic temperature.

Step-by-step reasoning

1. Identify the object whose temperature is measured. 2. Compute signed ΔT = T f − T i. 3. Select its total, specific or molar heat capacity with matching amount. 4. Calculate q object and include other apparatus contributions. 5. Reverse the sign to infer reaction heat under an insulated balance.

Visual explanation

Draw a reaction vessel inside a calorimeter bath. Show the thermometer rising from T i to T f and a heat arrow from reaction to bath. Write q bath = C bathΔT > 0 and q reaction ≈ −q bath beside opposite sides of the boundary.

Real-world analogy

A bucket and a cup can receive the same amount of warm water yet change temperature differently because their thermal capacities differ. Heat capacity tells how much energy is needed for a one-degree rise in the chosen object; the analogy does not replace the energy balance for a reaction.

Real-world example

If 100 g of water in a calorimeter warms by 2.0 K, a specific heat near 4.18 J g⁻¹ K⁻¹ gives about 836 J absorbed by the water. If the cup also absorbs heat, the reaction released more than 836 J in magnitude; its contribution must be added before inferring reaction heat.

Why?

Why can a large heat release cause only a small temperature rise? A large total heat capacity spreads the energy across substantial mass or apparatus, reducing ΔT for the same q.

Common misconception

“A 10 °C rise means 10 J of heat.” Temperature and energy have different units. The heat capacity and amount determine the energy corresponding to that rise.

Worked example

A calibrated calorimeter has total heat capacity 520 J K⁻¹. A reaction warms it from 298.0 K to 300.5 K, so ΔT = +2.5 K. The calorimeter absorbs q cal = 520 × 2.5 = +1300 J. If heat loss is negligible and the reacting material is treated as the complementary system, q rxn ≈ −1300 J for the amount reacted. To report kJ mol⁻¹, divide by the actual reaction extent in moles.

Quick check

1. A sample cools. What is the sign of its q = CΔT? Answer: Negative, because ΔT < 0 for cooling.

Exam focus

Name the body whose q is being calculated and reverse signs only when moving to the complementary reaction. Use matching heat-capacity units and account for apparatus heat capacity if supplied.

Advanced insight

Heat capacity is a temperature derivative of a state function under a specified constraint: C V relates to (∂U/∂T) V and C P to (∂H/∂T) P. This explains why one material can have different values under different mechanical conditions, especially in gases.

Summary

Heat capacity converts a temperature change into a heat estimate for a specified object. With nearly constant C, q = CΔT. Calorimetry uses the heat gained by solution and apparatus to infer the opposite heat of the reaction under a controlled energy balance.

Practice questions

1. A body with C = 200 J K⁻¹ warms by 3 K. Find q. Answer: +600 J for that body. 2. If the same body cools by 3 K, what is q? Answer: −600 J, because ΔT = −3 K. 3. A calorimeter gains 1.2 kJ from a reaction with negligible losses. What is q rxn? Answer: Approximately −1.2 kJ for the amount that reacted.