Experiment: The Mechanical Equivalent of Heat

Derivation. The observation refutes the hypothesis it was aimed at. Suppose heat is a conserved substance resident in bodies, as the caloric theory held. Then the heat obtainable from a given body is bounded by the amount it contains, and heat is released only by lowering that content—by a change of state, or by a reduction in the capacity of the substance to hold caloric. Both consequences are contradicted at once: the yield was unbounded and, so far as could be told, proportional to the work expended rather than to any property of the metal consumed; and the dust, which on the caloric account had given up its heat by a reduced capacity, showed the same capacity as the solid. Hence what is released is not drawn from a fixed store, and the conserved quantity—if there is one—must be one that mechanical work can supply. Identifying it is the work of Section 56.2; the caloric theory survives only as the special case in which no work is done.

Derivation. Take \(n\) moles. At constant volume the gas does no work, so all the heat supplied raises the internal energy: \(C_{V}=\dd U/\dd T\), using that \(U\) depends on the temperature alone—which is the content of Phenomenon 56.4 below, and the assumption Mayer's inference silently rests on. At constant pressure the same rise of temperature raises \(U\) by the same amount, since \(U\) depends on nothing else, but the gas additionally expands, by \(\Delta V=nR\Delta T/P\) from the equation of state \(PV=nRT\), and does work

The heat supplied at constant pressure therefore exceeds that supplied at constant volume by exactly \(nR\Delta T\), which is Equation (56.1). The inference now runs backwards. The left side of Equation (56.1) is a difference of two calorimetric measurements, expressed in whatever unit of heat the laboratory uses; the right side is a mechanical work, expressed in mechanical units. The equality between them is therefore a conversion factor between the two units, obtained without ever converting work into heat in an apparatus. Its weakness is equally plain: it is only as good as the assumption that a gas expanding freely does no internal work, which nothing in the argument tests. Joule's paddle wheel tests it.

Derivation. The work delivered in a run of \(N\) descents of two weights of mass \(M\) through a height \(h\) is

\(u\) being the residual speed of the weights at arrest and \(W_{\mathrm{f}}\) the work absorbed by the pulleys and the suspension, both of which Joule measured separately and subtracted. No heat crosses the insulated boundary, so by the first law the internal energy of the calorimeter rises by exactly \(W\). The calorimeter ends in a state differing from its initial one only in temperature, so that rise is \(C\,\Delta T\) with \(C\) the total heat capacity of water, vessel and fittings, and

Everything on the right of Equation (56.4) is a property of the calorimeter and of the work supplied; the stirring mechanism has dropped out. This is the whole content of the observation, and it is the empirical statement that the adiabatic work between two states depends only on the states—the property that makes the internal energy a state function in Classical Thermodynamics. Had the equivalent depended on the mechanism, one could raise the calorimeter's temperature by the cheaper route and lower it again by the dearer, returning every part of the apparatus to its initial state while extracting net work; the measurement is thus also a bound on perpetual motion of the first kind.

Derivation. The gas expands into vacuum, so it does no work on anything outside it, \(W=0\); the assembly is insulated, so \(Q=0\). By the first law \(\Delta U=0\): the free expansion is a change of volume at constant internal energy. The measurement reports \(\Delta T=0\) across it, that is

Expanding the derivative at constant \(U\) by the chain rule,

since \((\pp U/\pp T)_{V}=C_{V}\). The heat capacity is finite and positive, so Equation (56.5) forces \((\pp U/\pp V)_{T}=0\) and hence \(U=U(T)\). Together with the equation of state \(PV=nRT\) this completes the description of the ideal gas and yields Equation (56.1). The honest statement of the result is a bound rather than an identity: the experiment shows only that \((\pp U/\pp V)_{T}\) is smaller than the thermometry could resolve, the water bath's own heat capacity swamping the effect.

Derivation. Follow a fixed quantity of gas through the plug. Upstream, the piston pushing it in does work \(P_{1}V_{1}\) on it; downstream, the gas does work \(P_{2}V_{2}\) in making room for itself. The tube is insulated, so \(Q=0\), and the first law gives \(U_{2}-U_{1}=P_{1}V_{1}-P_{2}V_{2}\), that is

Throttling is a process at constant enthalpy—not a quasi-static one, but the end states are equilibrium states and that is all Equation (56.7) requires. The measured quantity is therefore

by the same chain rule as before with \((\pp H/\pp T)_{P}=C_{P}\). From \(\dd H=T\,\dd S+V\,\dd P\) and the Maxwell relation \((\pp S/\pp P)_{T}=-(\pp V/\pp T)_{P}\) of Classical Thermodynamics,

For an ideal gas \(V=nRT/P\) gives \(T(\pp V/\pp T)_{P}=V\) and \(\mu=0\): the effect vanishes identically, so whatever is measured is a departure from ideality and nothing else. The sign of Equation (56.9) changes where \(T(\pp V/\pp T)_{P}=V\), which is the inversion condition; for the van der Waals gas of Classical Thermodynamics the attraction \(a\) cools and the excluded volume \(b\) warms, and the inversion temperature is where the two balance—which is why hydrogen and helium, with small \(a\), must be precooled before throttling will liquefy them.