1886.] 



On the Specific Heats of Minerals. 



253 



each kind of molecule present in the substance; W the molecular 

 weight of the entire compound = 2 wn ; s the required specific heat 

 — the sp. h. of beryl is found to be 0*2140. The experimental re- 

 sults are 



a. Transparent crystals O2066 



/3. Clouded crystals 0-2126 



With the last variety the agreement is very close, but in either case 

 the agreement is sufficient to render the calculation strong evidence 

 had the case been one of hypothesis as to the nature of the sub- 

 stance. 



I proceed to quote a case where the identification of a mineral was 

 in this way left to the calorimeter. 



The mineral had been assigued a place in a small collection as 

 cryolite, on supposition. Its sp. h. was found to be 0*2558. This at 

 once distinguished it from gypsum — sp. h. = 0*278 — which it re- 

 sembled in appearance. As I knew of no experiments on cryolite it 

 was necessary to calculate the constant. The formula of cryolite — 

 3NaF + A1 2 F 3 — includes the sp. hs. of Na, F, and Al. Of these 

 F has not been directly determined. Its probable value was deduced 

 from the sp. h. of fluorspar, which I had found to be 0*2118. 

 Ascribing to the calcium in this compound the value 0*170 (Bunsen) T 

 the sp. h. of fluorine is deduced as 0*251, which agrees with Kopp's 

 result. The other atoms in cryolite were assumed as follows : to the 

 sodium was assigned Regnault's value 0*293, to the aluminium, that 

 obtained by myself, 0'223. On these assumptions the calculated 

 sp. h. of cryolite is 0*2569. The experiment on the hypothetical 

 cryolite affording 0"2558, the diagnosis was assured. A test of 

 fusibility now confirmed the result, and subsequently an authentic 

 specimen of cryolite gave 0*2538. 



The calculation may also be effected on the percentage composition. 

 It is a readier method but, it is remarkable, not so agreeable with the 

 practical result. Thus in the case of cryolite, Al = 13*21, F = 53*56, 

 Na = 32*65 in the mean analysis (Dana), and by the equation 



S x 100 = + w 2 s 2 , 



where w Y , w 2 , . . . are the percentages, s l5 s 2 , . . . the sp. hs. of the 

 constituents, the result is S = 0'2591. 



Tke specific heats of a few substances occasionally confounded 

 are gathered here. 



r Apatite, opaque 0*1920 



J „ translucent 0*1829 



] Beryl, clouded 0*2127 



I „ transparent. 0*2066 



