Page:Text-book of Electrochemistry.djvu/28

I. MECHANICAL WORK. 11 point of the scale, not the melting point of ice, but the "absolute zero," which lies 273° lower. If the temperature of a body is t° on the ordinary scale, then it is T° = 273° + t° on the absolute scale. T is called the "absolute temperature" of the body.

Mechanical Work. — The work which is done in raising a kilogram through 1 metre is a "kilogram-metre." In scientific measurements the unit of force is the "dyne," and is that force which the earth by its attraction exerts on $$\frac{1}{981} \text{gram}$$. Since the unit of length chosen is the centimetre =0.01 metre, the kilogram-metre (kg.m.) = $$9.81 \times 10^{7} \text{cm}$$. dynes = $$9.81 \times 10^{7} \text{ergs}$$; 1 erg = 1 cm. dyne, is the unit of work in the C.G.S. (centimetre, gram, second) system. Experimentally it has been determined that mechanical work of 426 gram-metres, or 0.426 kg.m. is required to produce one (small) calorie of heat. Consequently —

$$1 \text{cal.} = 9.81 \times 0.426 \times 10^{7} = 0.418 \times 10^{8} \text{ergs.}$$

In electricity the unity of work is the volt-coulomb, i.e. the work, which 1 coulomb balances over a fall of potential of 1 volt. For the value of this we have —

$$1 \text{volt-colomb} = \frac{1}{9.81} \text{kg.m.} = 0.24 \text{cal.}$$

Work done: Effect. — In the working of a machine we are concerned chiefly with the absolute value of the work done per second.

As a practical unit the horse-power has been chosen, which corresponds to a work of 75 kilog.met. per second. The electrical unit of work is the volt-ampere, or watt, which is equal to 1 volt-coulomb per second (since 1 amp. = 1 coulomb per second). As the watt is much too small a unit for measuring the work done in a dynamo, use is made of a unit 1000 times larger — the "kilowatt." It is quite evident that —

$$1 \text{kilowatt} = \frac{1000}{75 \times 9.81} = 1.36 \text{horse-power.}$$