408 DR. C. H. LEES ON THE THERMAL CONDUCTIVITIES OF SOLIDS 



The solution of this equation is 



/M/I . /)*'* 



10 'v x I-* "v x 



7?/i 



where A and B are constants of integration. 



Fig. 8. 



For the platinoid wire, w = '21/150 = '0014, p = '027, h = '0003, g = "002, 

 k = '08, therefore 



4 = 170 and A/4=' 22 ' 

 vh v gk 



Hence in the platinoid wire 



Vl - 170 = A,e~ + B,e-"'. 



For the copper wire, u< = '0014/40, p = '029, h = '0003, q = '004, k = 1, 

 therefore, 



= 4 and - -047. 



ph v j* 



Since the temperature of the copper wire to the right is nowhere infinite, we have 

 in the copper wire 



v, - 4 = B,e- 0471 



where B 2 is a constant of integration. 



When x = 1'6 in the expression for the temperature of the platinoid wire, the 

 temperature excess v { should be equal to that of the discs, which, on the average 

 was 17C. 



Hence, 



153 = A,e- 35 + B^' 35 = 705A, + 1'42B 1( 



Also at x = the temperatures and fluxes in the two wires are alike. 

 Therefore, 



170 + A, + B, = 4 + B 2 



and 35 (A, B,) = 190B 2 . 



These three equations determine A u B,, B 2 , and give for the temperature in the 

 platinoid wire, 



, - 170 = - 100e 22x - 57'8e- 22 *. 



The heat flowing per second from the wire to the discs at x = 1'6 



