Heat Capacities and Energy Storage

How much energy raises the temperature?

Heating a spoon and a bucket of water through the same temperature rise generally takes very different amounts of energy. The difference comes partly from the amount of material and partly from how that material stores energy. A third factor matters too: what is the material allowed to do while being heated? A gas that expands can transfer some input energy as work instead of retaining it internally.

Heat capacity S measures the heat supplied per unit temperature increase for the whole sample, under stated process conditions. Dividing S by sample mass m gives specific heat capacity s. Dividing by mole number n gives molar heat capacity C. A temperature interval of 1 °C has the same size as an interval of 1 K, but absolute gas temperatures still require kelvin. These definitions concern sensible heating without a phase change; during a phase transition heat can enter while ΔT = 0.

Do not identify Q/ΔT with a universal material constant without saying what is held fixed. A fixed-volume gas and the same gas under a freely moving loaded piston need different heat inputs for the same ΔT. Heat capacity is not a state function merely because it has a reproducible value under specified conditions. It quantifies a specified response to heating.

Why many solids have a molar heat capacity near 3R

A simple classical model pictures each atom in a solid vibrating about an equilibrium position. In one direction an oscillator has both kinetic and restoring-force potential energy. Classical equipartition assigns an average ½kBT to each quadratic energy contribution, so one direction contributes kBT in total. Three independent directions contribute 3kBT per atom. Here kB is Boltzmann’s constant and NA is Avogadro’s constant, with NAkB = R.

For one mole, U ≈ 3RT, hence an increase ΔT produces ΔU ≈ 3RΔT. At constant external pressure the first law gives Q = ΔU + PΔV. Since the volume change of an ordinary solid is small, PΔV is usually negligible in this estimate, so Q ≈ ΔU and C = Q/ΔT ≈ 3R. This is an approximate explanation, not an exact equality for every solid or every temperature.

Most listed molar values cluster near 24.9 J mol⁻¹ K⁻¹ even though the capacities per kilogram differ greatly. One kilogram contains different numbers of atoms for different elements, which explains part of that contrast. Carbon is a clear room-temperature exception. At low temperatures the classical prediction also fails generally: the full quantum explanation of vibrational excitation is outside this chapter. The table therefore tests the model and exposes its limits rather than proving a universal exact law.

Water, calorie and temperature dependence

The horizontal axis is water temperature in °C and the vertical axis is specific heat in cal g⁻¹ °C⁻¹. The curve falls from near 0 °C to a shallow minimum near 40 °C and then rises toward 100 °C. Its vertical scale is tightly expanded around 1: the variation is small, not a change by several multiples. A constant value near 4186 J kg⁻¹ K⁻¹ is useful for elementary calculations, while precision work requires temperature-dependent data.

Because water’s specific heat is not exactly constant, an older water-based calorie definition specifies an interval: the heat required to raise 1 g of water from 14.5 °C to 15.5 °C. For the chapter’s calculations, 1 cal ≈ 4.186 J, with rounded values such as 4.19 J used when explicitly supplied. The “mechanical equivalent of heat” is a conversion factor between energy units. Work and heat both use the joule in SI; they are not different kinds of substance.

Why Cₚ exceeds Cᵥ for an ideal gas

At constant volume a gas does no pressure–volume work. For one mole, the first law becomes QV = ΔU. Therefore Cᵥ = ΔU/ΔT. For an ideal gas U depends only on T at fixed amount and composition, so the same ΔT produces the same ΔU even on a different route. For n moles and approximately constant Cᵥ, ΔU = nCᵥΔT.

Now heat one mole through the same ΔT at constant pressure. It expands. The first law is QP = ΔU + PΔV. Divide by ΔT: Cₚ = ΔU/ΔT + PΔV/ΔT. The first term equals Cᵥ because U(T) is independent of the heating route. At fixed P, the one-mole ideal-gas equation PV = RT gives PΔV = RΔT. The second term is therefore R.

The extra heat needed at constant pressure does not disappear. For each mole and each kelvin of temperature rise, R joules are transferred as expansion work. Both processes produce the same internal-energy rise. Cₚ − Cᵥ = R is an ideal-gas relation; it is not an exact formula for liquids, solids or arbitrary real gases. The ratio γ is dimensionless and will determine the shape of reversible adiabatic curves.

What high specific heat changes in everyday life

When two insulated bodies exchange energy without phase change or appreciable work, energy conservation gives m₁s₁(Tf − T₁) + m₂s₂(Tf − T₂) = 0. If the capacities are approximately constant, Tf is a heat-capacity-weighted mean. The arithmetic mean occurs only when the two total heat capacities are equal, not whenever two bodies touch.

A coolant with high specific heat can absorb much energy with a modest temperature increase, helping prevent overheating. Water’s high heat capacity also reduces the temperature swing of a large body of water for a given heat input. Coastal locations therefore often have more moderate temperature variation than dry inland areas at similar latitude. Water movement, winds and evaporation contribute too; heat capacity is a major part of the thermal-inertia explanation, not the only climate factor.

A car tyre warms during driving because mechanical deformation and frictional processes dissipate energy. With approximately fixed gas amount and tyre volume, P/T is approximately constant, so warming raises the absolute gas pressure. In a real tyre, slight volume changes and heat transfer also occur. The correct explanation distinguishes the cause of warming from the equation that connects the warmer temperature to pressure.

Quick Quiz