442 Specific Heat of the different Gases. [Dec. 



we should have S — V T : but it is not given immediately by expe- 

 riment. We must therefore endeavour to determine it by joining, 

 to the data which we have already, that which we obtain by the 

 knowledge of the space which the body setting out from the point, 

 C, passes over with a variable velocity, v, which at first is equal to 

 V. This space is given us by experiment. We shall express it by 

 A — B : A being the known distance, C D, and B the space that ■ 

 remains to be passed over at the end of the time, T. For the 

 greater convenience, let us suppose for a moment that it is the 

 space A — B that we have to determine, B being unknown, and 

 V being known. Its expression will be d (A — x), or — dx = 

 v dt, x being the general expression of the space which remains to 

 be passed over each instant, v that of the velocity at the same time, 

 and t that of the time elapsed. But since at each instant the 

 velocity is proportional to the space that remains to be passed over, 

 we may make » = k x, k being a constant quantity which may be 

 determined by making k A = V. We shall have then — d x = 



Jcxdt, and = idt. Integrating the sides of this equation, 



we obtain — log. hyp. ,r -f C = k t. To determine the con- 

 stant quantity, C, make x = A, which suppose the time, t, to be 

 nothing. We get — log. A + C = o, and C = log. hyp. A. 

 Substituting this value of C in the preceding equation, we shall 



— • A A 



have — log. x -f log. A, or log. — = k t, — , = e u (e being the 



number whose hyperbolic logarithi* is unity ;) x = -77 , and A — 



x = A — -jg- . Applying this equation to the time given, T, x 



will become B, and we shall have B = -g , and A — B = A — 

 A 



Let us suppose, now, that B and A — B are given by experi- 

 ment, and that V, or, which comes to the same thing, k, is unknown. 

 To discover its value, it is sufficient to make use of the equation, 

 which on the preceding hypothesis gave the value of A — B, sup- 

 posing B known ; or still more simply to employ the equation log. 



A 



— = k t, to which we came by putting for x and t their values B 



, A A 



and T. We obtain from it i = g ' B , V = A ° g ' B , and S 



T T 



= A log. — . We have seen above that S was the space passed 



over by the body with a uniform velocity, V, during the time, T. 



To apply this formula to the cooling of a hot body, we must 

 denote by A the excess of its temperature above that of the air at 

 the beginning of the experiment ; and by B, the same excess at the 

 end of the time, T. 



