Julius Robert Mayer and the principle of energy conservation

Abstract

Julius Robert von Mayer was among the first to state the principle of energy conservation. As a medical doctor, he was interested in the relationship between heat and work, being stimulated by observations and considerations regarding the physiology of the human body. In 1842, he published a first manuscript in which he reported the value of the mechanical equivalent of heat; then, in a paper in 1845, he elucidated the procedure followed for the calculation of the equivalent. The result achieved by Mayer, based only on theoretical calculations, was presented before the experimentally determined value obtained by Joule in 1847 and remains of capital interest in the history of thermodynamics.

Appendix: Calculation of the mechanical equivalent of heat

Mayer calculates the “mechanical effect” y (that we can trace back to the meaning of work) related to the heating at the constant pressure of 1 cm³ of air from 0 °C to 1 °C. Taking into account a column of mercury, with base 1 cm² and height 76 cm, the volume of the column is:

$$V_{{{\text{Hg}}}} = {\text{ 1 cm}}^{{\text{2}}}\, {\text{76 cm }} = {\text{ 76 cm}}^{{\text{3}}} .$$

Knowing the density of mercury at 0 °C (d_Hg = 13.60 g cm⁻³), the weight (indeed, the mass) of the column, W_Hg, is:

$$W_{{{\text{Hg}}}} = V_{{{\text{Hg}}}} {\text{d}}_{{{\text{Hg}}}} = {\text{ 76 cm}}^{{\text{3}}} {\text{13}}.{\text{6}}0{\text{ g cm}}^{{ - {\text{3}}}} = {\text{ 1}}0{\text{33 g}}.$$

The volume variation (ΔV) associated to the expansion of 1 cm³ of air heated from 0 °C to 1 °C at the constant pressure of 76 cm of Hg is: ΔV = (1/274) cm³. Thus, considering the base section of the column equal to 1 cm², the raising of mercury in the column is: h = 1/274 cm = 0.00365 cm. Consequently, the mechanical effect y due to the air expansion results:

$$y = Wh = {\text{ 1}}0{\text{33 g }}\left( {{\text{1}}/{\text{274}}} \right){\text{ cm }} = {\text{ 1}}0{\text{33 g }}0.00{\text{365}}\;{\text{cm }} = 3.770{\text{ g cm}}.$$

Then, Mayer calculates the heat associated to the heating of 1 cm³ of air from 0 °C to 1 °C at constant pressure. The specific heat of air warming at constant pressure (C_P = 0.267 cal g⁻¹ °C⁻¹) was known; moreover, also known was the value of the ratio between C_P and C_V, the specific heat of air warming at constant volume:

$$C_{{\text{P}}} /C_{{\text{V}}} = {\text{ 1}}.{\text{421}}.$$

From these values results:

$$C_{{\text{V}}} = 0.188{\text{ cal g}}^{{ - 1}} \;^\circ {\text{C}}^{{ - 1}} .$$

The weight (indeed, the mass) of 1 cm³ of air at 0 °C and at the pressure of 76 cm of Hg can be calculated, knowing the air density at 0 °C. It results as: W_air = 0.0013 g. The heat adsorbed at constant pressure, Q_P, for increasing the temperature by 1 °C is:

$$Q_{{\text{P}}} = W_{{{\text{air}}}} C_{{\text{P}}} \Delta T = 0.0013{\text{ g}}~\;0.267{\text{ cal g}}^{{ - 1}} \;^{ \circ } {\text{C}}^{{ - 1}} 1\;^{ \circ } {\text{C}} = 0.000347{\text{ cal}}.$$

whereas the heat adsorbed at constant volume, Q_V, is:

$$Q_{{\text{V}}} = W_{{{\text{air}}}} C_{{\text{V}}} \Delta T = 0.0013{\text{ g }}0.188{\text{ cal g}}^{{ - 1}} \;^{ \circ } {\text{C}}^{{ - 1}} 1\;^{ \circ } {\text{C }} = 0.000244\;{\text{cal}}.$$

The difference:

$$Q_{{\text{P}}} {-}Q_{{\text{V}}} = 0.000{\text{347 }}-0.000{\text{244 }} = 0.00010{\text{3 cal}}.$$

corresponds to the ability of the system to lift a weight of 1033 g by 1/274 cm and is related to the previously determined mechanical effect y.

From these values, Mayer can now calculate the proportionality constant k, i.e. the mechanical equivalent of the heat, from the ratio:

$$k = y/\left( {Q_{{\text{P}}} {-}Q_{{\text{V}}} } \right) = \left( {{\text{3}}.{\text{77}}0{\text{ g cm}}/0.000{\text{1}}0{\text{3 cal}}} \right) = {\text{36,7}}00{\text{ g cm cal}}^{{ - {\text{1}}}} = {\text{ 367}}\;{\text{g m cal}}^{{ - 1}} .$$

Finally, taking into account the transformation of the mass in weight and converting the result to the International System of Units, we obtain:

$$k = \left( {{\text{367 g m cal}}^{{ - {\text{1}}}} {\text{9}}.{\text{8 m s}}^{{ - {\text{2}}}} } \right)/\left( {{\text{1}}000{\text{ g kg}}^{{ - {\text{1}}}} } \right) = {\text{3}}.{\text{6}}0{\text{ kg m}}^{{\text{2}}} {\text{s}}^{{ - {\text{2}}}}/{\text{cal }} = 3.60{\text{ J/cal}}.$$