436 



HYDRODYNAMICS. 



By adding to each of these results its product by the dilatation *= ^TO in mercury, we have 



=1342,66812+3,101 13=1345,7692 Exp. 1. 



Specific 

 Gravities. 



QQ 11+^=1340,28636+5,10160=1345,3880 Exp. 2. 

 1 + K. { 



The arithmetical mean between these results is 

 1345,5786 grammes, which, being divided by V, al- 

 ready found to be 98,960036, we have 



When the 

 body is so- 

 luble in 

 water. 



When the 

 body im- 

 bibes wa- 

 ter. 



'-98,965036 



which is the weight of a cubic centimetre of mercury 

 in grammes, at the temperature of melting ice. 



If we wish to compare this weight with that of wa- 

 ter, we have only to calculate the last for the tempe- 

 rature of melting ice, or for 3.42 of the centigrade 

 thermometer. But if J is the dilatation of water from 

 its maximum density to its freezing point, or for 



3.42, the weight of water required will be - r, and 



the relation between the weights of mercury and water 

 at the temperature of melting ice will be jr (1+3). 

 But, by substituting 3.42 in the formula of dilata- 

 tion, No. 1. we shall have 2= 0,0000746. Hence 



(1+3)=13,597I90+0,001017=13,598207, 



which is the exact ratio between the weights of equal 

 volumes of mercury and water at the temperature of 

 melting ice. 



PROP. XIII. PROB. 



To determine accurately the specific gravities of so- 

 lid bodies. 



In determining the specific gravities of solid bodies, 

 we may adopt two methods. 1. We may weigh them 

 successively in air and in some other fluid, which is 

 the ordinary method ; and then, if P is the apparent 

 weight of the solid in air, and p the weight of the volume 



of water which it displaces, we have for the specific 



gravity of the body, neglecting the necessary reduc- 

 tions ; or, 2. After having weighed the solid in air, we 

 may place the solid in a glass vessel, and weigh them 

 -conjointly, and then weigh the same glass vessel when 

 filled only with water. If then D is the weight of the 

 solid in air, p the weight of the vessel containing the 

 water and the solid, and p' the weight of the vessel con- 

 taining water alone ; then p p' is the weight of the 



quantity of water displaced, and =5 -- -. will be the 



1 P+P 

 specific gravity required. 



If the substance is soluble in water, like many of the 

 salts, it is necessary to use alcohol, or some other fluid, 

 such as the essential or fat oils, which are not capable 

 of dissolving it. The specific gravity of the oil being 

 known, that of the salt will be immediately found. 



If the solid imbibes water, without either dissolving 

 or decomposing it, it is necessary first to weigh the 

 body when perfectly dry, which weight we may call P, 

 and then weigh it when it has imbibed as much water 

 as possible. Let this weight be P'. We must next 

 find how much water the body displaces, which we may 

 call a, then the apparent specific gravity of the body is 



, as it has really displaced the quantity of water c. 



But, in order to know the specific gravity of the solid 

 parts of the body which do not admit water, such as 

 the real fibrous part of sponges, then we must con- 

 sider, that the quantity of water displaced is not merely 



a, but a P', and therefore is the real specific gra- 



vity, neglecting the necessary reductions. 



In order to explain the formulae given by M. Biot for 

 solid bodies, let us take 



t = temperature at which the solid is weighed. 



V = the volume of the solid body in cubic centimetres 



at the temperature I. 

 (s) 55 the absolute weight of a cubic centimetre of its 



substance at the temperature of melting ice. 

 K = the cubical dilatation of , the solid^for one degree 



of the centigrade thermometer. 

 (e) = the weight of a cubical centimetre of water at 



the temperature of melting ice. 

 J = the dilatation of water from 32 to /. 

 a. = the ratio of the weight of air to that of water in 



the circumstances under which the experiment 



is made. 

 A = the dilatation of any other liquid employed in- 



stead of water. 

 (51-)== the weight of a cubical centimetre of another li- 



quid at the temperature of melting ice. 

 S r= the weight of the solid in air. 

 S' = the weight of the solid in water. 



Case 1. When the body is weighed successively i n 

 air and water, 



S-S' 



If the body has been weighed successively in water 

 and in air at the same temperature, then (a-)rr(e) and 

 A=3, consequently 



,.N_ 



no.^ rs SM No - 2 - 



(1 +oj ^o o 1 



Case 2. When there are three weighings, 1 st of the 

 solid body, 2d of the glass in a vessel filled with a liquid, 

 and 3d of the same vessel containing the solid and the 

 liquid ; 



S P+L 

 If the body has been weighed successively in water 



and in air, then (5r) = ( 

 reduced to 



, . 



and ArrJ, and the formula is 



N 



Case 3. When there are only two weighings, 1st of 

 the solid in air, and 2d of the solid in liquid in the 

 same vessel : In this case let M be the weight of the so- 

 lid and liquid, 



