Degrees of Heat in Absolute Measure. 75 



In combination with equation (10) these equations give 



dk -j— dk — dk -r- 



* + * + _* a . . . (11) 



ax ay dz 



"VVe see, therefore, that in the motion of heat which we have 

 here imagined, in which the suppositions are quite analogous to 

 the conditions which actually obtain for the permanent motion 

 of electricity, the temperature would be determinable by the 

 same differential equation as the electrical tension. 



By the permanent motion of electricity through a body heat 

 results, which ultimately acquires a permanent motion when it 

 is continuously conducted away in the same manner. The in- 

 crease of energy which owes its origin bolh to the motion of elec- 

 tricity and to that of heat is now nil in each element of the 

 body ; and from the equations (2) and (7) we shall obtain 



d /c i(F + T- 2 ) d k d{?* + T> ) t d Jt rf(P*+T« ) =a 

 dx dx dy dy dz dz 



By this equation, in combination with equation (8), both the 

 electrical tension P and the absolute temperature T are to be 

 determined when both the motion of electricity and that of heat 

 are become permanent. 



If, for instance, we pass electricity through a body, keep- 

 ing in a small part cr Q of its surface a constant electrical tension 

 P , and in another part <r l of its surface the tension P T , and if at 

 the same time we maintain both surfaces at the same constant 

 temperature T , while the other part of its surface is kept sur- 

 rounded by perfectly bad conductors of heat and electricity, a 

 permanent motion of electricity and of heat will ultimately result, 

 and the electricity will develop the same heat as is conducted 

 away by the surfaces cr and a v 



If we put 



P* + T 2 + AP + B = <k (13) 



in which A and B are two arbitrary constants, we shall obtain 

 by the equations (12) and (8), 



d k d± + £ k dj> + d k dj L=(i _ _ _ 



dx dx ay ay dz dz 



Both constants A and B will then be so determined that for 

 both surfaces a and <r l <£ = 0, by putting 



PJ + T^ + AP + B = 0, 



P? + T2 + AP 1 + B = 0. 



With the values of A and B which result from these, namely 



A=-(P + P 1 )aadB = P P 1 -T 2 , . . . (15) 



