Now

made no assumptions as to the nature of the working substance. if the efficiencies of a number of reversible engines with different working substances are all the maximum, they must all be equal.

The above result may not appear to agree with common sense, for if the working substance in an engine is ether, say, we might at first sight hope to obtain a higher efficiency than with water. For the vapour pressure of ether at any temperature being higher than that of water at the same temperature, the pressure in a boiler filled with ether would be greater than in a boiler filled with water at the same temperature. Thus the vapour supplied to the ether engine will be at a higher pressure than that supplied by the water engine, and so we might expect that we should get more work by allowing the ether vapour to force back a piston than in the case of the steam. It must, however, be remembered that at the back of the piston we have acting the pressure of the exhaust steam or vapour, and that the condenser for the two engines must be supposed to be at the same temperature, or they would not be working through the same range. Now the vapour pressure of the ether at the temperature of the condenser will be greater than that of the water, and so not only the forward pressure in the ether engine but also the back pressure is greater than in the water engine, and so no advantage is gained.

Hence a Carnot's cycle being a thermal process which is independent of the nature of the substance in which the thermal changes take place, it at once becomes of interest to see whether we cannot utilise this fact in order to define a scale of temperature independent of all properties of any particular kind of matter. The scales which we have used heretofore all depend on the change of some one physical property of some special kind of matter, thus on the increase in volume of mercury or hydrogen, the increase in resistance of a platinum wire, the thermo-electromotive force of a junction of two given metals, &c. A scale of temperature depending on Carnot's cycle and independent of the properties of any particular kind of matter has, however, been devised by Lord Kelvin, and to this scale only can the title "absolute" be given with justice.

If, as before, we imagine a Carnot's cycle in which a quantity of heat H1 is drawn from a source at a temperature 71, and an amount of work Wis performed, H, units of heat being given out to the refrigerator at the temperature 7, we may according to the first law measure H, and H1 in terms of ergs, in which case H1- H2=W. If now, keeping H1 and T1 constant, we adjust the temperature 7, so that the work done during the cycle is unity, then the two temperatures 72 and 7, will be such that if a Carnot's engine working between these temperatures takes H1 ergs of heat from the source it will perform one erg of work. Next suppose that another cycle is taken, in which the lower temperature T is so adjusted that when H, ergs of heat are drawn from the source at a temperature T1, the work done in the cycle is two ergs. Then, according to Lord Kelvin, the difference of temperature between 7, and T is to be

Pro

called twice the difference of temperature between 7, and T ceeding in this way, we could define a series of equal temperature intervals, and thus form a thermometric scale. It will, however, be convenient not to call the interval T1- T2, or 72-73, as above defined, one degree, since the scale thus constructed would not resemble the scale ordinarily employed. We will therefore suppose that 7, is taken as the temperature of boiling water, and we will postulate that when H1 units of heat are taken, by an engine working in a simple reversible cycle, from a source at the temperature of boiling water, and the refrigerator is at the temperature of melting ice, a hundred times the work will be done that would be done supposing the temperature of the refrigerator were one degree, on this new absolute scale, below the temperature of boiling water, and so on.

Let the lines ToTo and T10T10 (Fig. 212) be the isothermals for the temperatures of

melting ice and boiling water respectively, and let AB be an adiabatic cutting these isothermals at E and G. Suppose that if we go along the isothermal T10 from

E to F an amount of heat H10 (measured in ergs) has to be supplied to the working substance to keep its temperature constant, and that through F we draw a second adiabatic CD cutting the isothermal To at H. Then, if a simple reversible engine performs the cycle EFGH, it will take in Ho units of heat at a temperature T10 and give out

Ho units of heat at a temperature To, while the work W10 done during

the cycle will be represented by the area of the figure EFHG.

Now draw

nine isothermals between T, and Te so spaced that the area intercepted between any adjacent two and the two adiabatics is one-tenth of the area EFGH. Thus the area shown shaded is to be one-tenth of EFGH. Then the temperatures corresponding to these isothermals, if we call the temperature of melting ice 273° and that of boiling water 373°, are 283°, 293 ̊, 393 ̊, 313 ̊, 323", 333°, 343°, 353, 363° on Lord Kelvin's absolute scale. By the doctrine of the conservation of energy, the maximum amount of work we can possibly get from a quantity of heat H1 is JH1, if the quantity H, is expressed in calories, or simply H1, if this quantity is expressed in ergs. Keeping the temperature T1 of the source constant, the amount of work W obtained during a cycle will increase as the temperature of the refrigerator is lowered, until the temperature of the refrigerator becomes such that no heat is given to it during the compression portion of the cycle, the whole of the heat taken in being converted into work, so that H1=W, or the efficiency of the cycle becomes unity. We cannot imagine a refrigerator at a lower temperature than this, and hence may take it as the zero on this new absolute scale. It is found that the zero thus defined coincides with the absolute zero as given by a perfect gas, and that the new absolute scale agrees very nearly with that of a gas thermometer containing a perfect gas. So that the use of the thermometric scale derived from the expansion of a perfect gas is justified.

If, in Fig. 212, any other adiabatic KL is drawn, then this, together with either of the others, will cut off equal areas between consecutive isothermals. Thus the area intercepted by any two adiabatics and any two isothermals 7, and 72, say, will be proportional to the difference of temperature 7 - T, for each degree of this difference will correspond to an equal small area k, such as the one shown shaded. Thus we shall have W= k(T1 − T2),

where is a constant depending on the two adiabatics taken. WH, H, this gives

H1- H2=k(T1- T1).

Since

Now if we make 7, the absolute zero, there will be 7, small areas each equal to & included in the cycle, and H, will be zero, so that in this case

W'= H1 = kT1

Now the efficiency of a reversible cycle is given by

This result will be found useful when we are considering some actual cases of reversible cycles.

The above equation may be written

or the ratio of the heat taken from the source by a reversible engine to the heat given up to the refrigerator is the same as the ratio of the temperature, on the absolute scale, of the source to that of the refrigerator.

262*. The Second Law of Thermo-Dynamics.-When considering the efficiency of a simple reversible engine, we said that the transfer of the heat of the condenser to the source was contrary to experience. The denial of the possibility of any such action forms what is called the second law of thermo-dynamics,1 and has been put into a concise form by Clausius, who expresses it as follows: It is impossible for a self-acting machine, unaided by any external agency, to convey heat from one body to another at a higher temperature.

Lord Kelvin has enunciated the second law in a slightly different form, namely: It is impossible, by means of inanimate material agency, to derive mechanical effect from any portion of matter by cooling it below the temperature of the coldest of surrounding bodies.

It must be carefully borne in mind that these laws refer only to the work performed during a cycle of operations, in which the initial and final states of the working substance are exactly the same. Thus when a gas is allowed to expand against external pressure, it does work and becomes cooled, so that in this way it may do work although in the operation it becomes cooled below the temperature of surrounding objects. The final state of the gas is not, however, the same as the initial state, and if we attempt to bring the gas back into the initial state we shall find that the law holds.

We may also put the law in slightly different words, viz. that heat of itself never passes from one body to another at a higher temperature ; and if by any means we cause heat to be transferred from a body to another at a higher temperature, we must in the process supply the system with energy from some outside source. Thus, when a reversible engine is worked backwards, heat is taken from the refrigerator and supplied to the source. During this operation, however, external energy has to be supplied to the engine, so that it is not working "by itself."

263*. Calculation of the Effect of an Increase of Pressure on the Melting Point of Ice.--The second law of thermo-dynamics will allow us to calculate the effect of pressure on the melting-point of ice.

1 The deductions made in the last section are also generally referred to as forming part of the second law.

PRESSURE

A

B

Suppose we have a gram of water at o° C. and at a pressure of one atmosphere, the conditions being represented by the point A, Fig. 213. Now allow the water to freeze. During this process the temperature and pressure will remain constant, so that the horizontal line AB will represent the change, which is an isothermal one. During this change 80 calories of heat, or, if we use mechanical units, 80 × ergs will be given out, and an amount of work pa̸', where v' is the change of volume, will be done. One atmosphere being 1013300 dynes per square centimetre,

D

VOLUME

FIG. 213.

C

and being in the case before us 0.0907, the work done is

1013300 X 0.0907 ergs.

Now, without allowing heat to enter or leave the ice, reduce the pressure to zero. Since the change in volume of ice (or water) with a difference of one atmosphere is quite inappreciable, the line BC showing this change is vertical, and no work is done, supposing that during the change the ice does not melt, that is, if the melting-point of ice does not vary with the temperature. Next, supply heat to the ice so that it melts, and we now pass along CD. During this process heat is absorbed, but since the pressure is zero, although the volume decreases, no external work is done on the working substance. Finally, raise the pressure to one atmosphere along the adiabatic DA. We have now gone through a reversible cycle of operations in which an amount of work represented by the area ABCD has been done. On the supposition, however, that the melting-point of ice is the same at a pressure of one atmosphere as in a vacuum, the temperature at which the heat was taken in is the same as that at which the heat was given out to the refrigerator. This is, however, contrary to the second law of thermo-dynamics, and hence we conclude that our supposition that the melting-point of ice is unaltered by change of pressure must be wrong. Since external work is done during the cycle, the temperature when the heat was being taken in by the working substance, that is, while the ice was melting at the low pressure along CD, must have been higher than the temperature when the heat was being given out, that is, when the water was freezing at the higher pressure. In short, decreasing the pressure has raised the melting-point.

We may proceed to calculate what would be the rise in the meltingpoint produced by an increase of one atmosphere. Let be the difference in the temperature of melting ice produced by a change in pressure of one atmosphere. The temperature of the refrigerator is o° C., or 273° on the absolute scale, and that of the source 273+. The heat absorbed is 80 calories or 3.352 × 10 ergs, and the work done is 0.0907p ergs,