W. P. White — Melting Point Determination. 463 



rlB 

 ture gradient, that is, as equal to 47rr 2 K — ; and (2) as the 



heat absorbed by the inclosed volume, or the product of volume, 

 volume specific heat, and the rate of temperature rise, 



, 47rr 3 n JO 



equal to lb — • 



1 3 dt 



dB 

 Equating these two and integrating, remembering that — 



Civ 



is here a constant, we have 



If the charge is an infinitely long cylinder, we have the same 

 expression with 4 substituted for 6 in the denominator of the 

 second member. In general, we may say, therefore, that the 

 temperature gradient within a uniform, solid, steadily heated 

 charge approximates a parabola, and the difference of tem- 

 perature between center and outside for bodies of the same 

 shape is proportional to the first power of the rate of heating 

 and to the square of the diameter. 



If we let 80 = A#, then St is the time required for a given 

 temperature value to pass from H to /\ — that is, it is the time 

 lag of r behind R. It equals 



It is independent of the rate and is thus important as a prop- 

 erty of the charge alone. 



2. Effect of the melting on the temperature distribution. — 

 In an actual melting point determination one fundamental 

 hypothesis of the preceding no longer holds. The tempera- 

 ture rise of the charge is not regular after melting begins. We 

 may, however, still proceed for the case of a small charge, for 

 which the time required to reach an equilibrium temperature 

 distribution will be relatively small, so that the divergence 

 from such a distribution may be neglected in making the first 

 approximation. Mathematically, the problem then reduces 

 simply to finding the effect of a change in the specific heat, 



the rate of heat supply, -=— , remaining constant. This is easily 



done as follows : Since 



dQ = Sd$ 



dO__dQ_ 1_ 



dt ~ dt ' S { ' } 



