192 



R. Clausius on the Application of the 



This sum must be equated to zero, in consequence of the condi- 

 tion now laid down that heat must neither be communicated to 

 nor taken from the mass. In this way we obtain, if we simply 

 write dm for 



dm dm irn 



-d^ dv +df dT ' 



the equation 



(13) rdm + m(h-c)dT+McdT=0. 



If we substitute in this, according to (12) 



dr r 



dr 



and again write simply dr for ——dT, since r is only a function 



a j. 



of T, we have 



rdm~^mdr-^dT+McdT=0 1 



777 Y 



or (14) d(mr)-~ dT+McdT=0. 



If we divide this equation by T, and remember that 

 d(mr) mr jf mr \ 



we obtain 



/ \ y /mr\ , dT 



(15) d(- f -j+Mc Y =0. 



As the specific heat of a liquid changes but slowly with the 

 temperature, we will in what follows, always consider the quan- 

 tity c as constant. Then the previous question may be inte- 

 grated at once, and gives 



— -f- Mc log jT=const. 



or if the initial values of T, r, m, be denoted by Ti, n, mi, 

 mr m r T 



( vn ) ~T r ~~ 1 r~^ ~~ g Y t ' 



By this equation, m is also determined as a function of the tem- 

 perature, if r, as a function of the temperature, can be a priori 

 considered as known. 



In order to give an approximate view of the behavior of this 

 function, I have collected together in the following table some 

 values calculated for a particular case. It is assumed namely 

 that the vessel at the beginning contains no liquid water, but is 

 exactly filled with steam at the maximum density, so that in 

 the previous equation m, is to be put equal to M, and let now 

 an expansion of the vessel take place. If the vessel should be 



