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Heat enters only during the constant-pressure step, so the heat capacity at constant pressure givesThe positive magnitude of the heat rejected during constant-volume cooling isBecause the working gas returns to its initial state, the thermodynamic cycle has zero net change in internal energy and $W=Q_1-Q_2$. Its thermal efficiency is therefore

The required Legendre transform is the Helmholtz free energy, $\boxed{F=U-TS}$ up to an additive constant. Its differential is $dF=dU-TdS-SdT=-SdT-pdV$, so $F_T{|}_V=-S$ and $F_V{|}_T=-p$. Equality of mixed partial derivatives for a smooth thermodynamic state gives the Maxwell relationNext use the total differential $dS=S_T{|}_{V}dT+S_V{|}_{T}dV$. Along a fixed-pressure path, divide by $dT$ to get $S_T{|}_p=S_T{|}_V+S_V{|}_T{V}_T{|}_p$. Substitute the Maxwell relation to obtainMultiplying by $T$ gives the usual difference between the heat capacity at constant pressure and the heat capacity at constant volume.

Temperature describes thermal state and determines the direction of spontaneous heat transfer. A quantity of heat is energy transferred because of a temperature difference. It depends on the process and is measured in energy units; it is not another name for temperature. Equal heat inputs need not produce equal temperature rises in different masses or substances.

For a specified process, the specific heat capacity isOver a modest range with nearly constant $c$ and no phase change, $Q=mc\Delta T$. For a gas the constraint matters: heat capacity at constant pressure and heat capacity at constant volume generally differ because expansion can do work.

In method-of-mixtures calorimetry, a body of known mass $m_s$ at $T_s$ is transferred into water of mass $m_w$ at $T_w$ in a calibrated calorimeter initially also at $T_w$. If the final uniform temperature is $T_f$, the vessel and immersed fittings have heat capacity $C_{\mathrm{cal}}$, and losses are negligible,ThusStirring, quick transfer, a lid, a temperature-time record and allowance for surrounding heat exchange improve the inference. Ignoring the calorimeter's heat capacity systematically misattributes part of the energy uptake.

An alternative is electrical calorimetry: measure $E=\int VIdt$ supplied to a heater and compare the temperature rise with a blank or separately calibrated vessel. At nearly constant properties, $E=(mc+C_{\mathrm{cal}})\Delta T+Q_{\mathrm{loss}}$. An ice calorimeter instead measures the mass melted at a known phase-change temperature, using $Q=m_{\mathrm{melt}}{L}_f$ with appropriate corrections. Steam condensation similarly uses latent heat. These methods infer transferred energy from different measurable effects; a thermometer reading alone does not measure a quantity of heat.

The Debye model treats a solid's low-frequency lattice vibrations as acoustic phonons in an elastic continuum. There are three polarizations per wavevector, one longitudinal and two transverse, with approximately linear dispersion relations. A cutoff is chosen to preserve the total $3N$ vibrational modes of $N$ atoms. The phonon number itself is not fixed; mode occupations follow the Bose-Einstein distribution with zero chemical potential.

For one polarization of speed $v_s$, counting wavevector states in a spherical shell gives $g_s(\omega )=V{\omega }^2/(2{\pi }^2{v}_s^3)$. Define an average Debye speed by $3/v_D^3=1/v_L^3+2/v_T^3$. The total density of states can then be writtenThe Debye wavevector is $k_D=(6{\pi }^{2}N/V{)}^{1/3}$, the Debye frequency is ${\omega }_D=v_D{k}_D$, and the Debye temperature is ${\Theta }_D=\hslash {\omega }_D/k_B$.

Omitting the temperature-independent zero-point energy, the vibrational energy and resulting Debye heat capacity areAt low temperature the upper limit can be extended to infinity. The integral is $4{\pi }^4/15$, givingOnly the long-wavelength acoustic modes are appreciably populated, explaining the Debye T-cubed test. At high temperature every mode contributes $k_B$ by equipartition theorem, so $\boxed{C_V\to 3N{k}_B}$, the Dulong-Petit law.

The model succeeds especially at low temperature, where long-wavelength elasticity is reliable. It replaces real anisotropic branches by a simple continuum and cannot reproduce detailed optical phonon branches or every intermediate-temperature feature. Metals also have an electronic heat capacity approximately proportional to $T$; defects, magnetic excitations or a phase transition can add other terms. Experiments often measure heat capacity at constant pressure, which need not equal the harmonic model's $C_V$ exactly.

Assume sodium supplies one conduction electron per atom. The supplied Fermi energy givesThe gas is strongly degenerate at room temperature. For one mole, $N=N_A$ and $N_A{k}_B=R=8.31446\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. Using the requested estimate,The supplied mass density determines the answer per unit volume: $N/V=(970/0.023)N_A=2.54\times 10^{28}{\mathrm{m}}^{-3}$. ThereforeThe density is not needed for the molar heat capacity once $E_F$ is supplied. Using the precise Sommerfeld expansion coefficient instead gives $0.327\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$.

Either electronic estimate is only about $1\%$ of the stated measured total. Most of the total heat capacity comes from lattice phonons, whose high-temperature Dulong-Petit law limit is $3R=24.94\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. The remaining difference from $28\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ need not be electronic: anharmonic lattice effects and the difference between heat capacity at constant pressure and heat capacity at constant volume also contribute. The paper does not specify the measurement constraint, so the two totals should not be equated exactly.

This is the Brayton cycle. Assume fixed gas amount, constant heat capacities, and internally reversible adiabatic processes, as intended by the ideal engine model. Put $r_p=p_h/p_l=p_2/p_1>1$ andCombining $p{V}^{\gamma }=\text{constant}$ with the ideal gas law gives $T\propto p^{(\gamma -1)/\gamma }$ along either reversible ideal-gas adiabat. ThereforeDuring a constant-pressure process the heat supplied equals the enthalpy increase. If $C_p$ is the molar heat capacity at constant pressure, the positive heat input and output areThe insulated branches exchange no heat, so the first law of thermodynamics gives $W=Q_{\mathrm{in}}-Q_{\mathrm{out}}$. HenceAn adiabatic process alone need not be an isentropic process; the relations used here also require internal reversibility. Dissipative compression or expansion would require extra information and would not, in general, obey this efficiency formula.

The specific heat capacity at fixed pressure and composition is $c_P=(\partial h/\partial T{)}_P$, where $h$ is specific enthalpy. Equivalently, it is heat capacity at constant pressure divided by mass. Under the process assumptions in heat capacity, it includes expansion work as well as the internal energy change. The sensible heat example uses this response separately for ice and water.