GIBBS' PAPERS I AND II 21 



temperatures. If you have two engines both using the same 

 amount of heat, they must do the same amount of work. For 

 if they do not, running one direct and the other reversed will do 

 a net amount of work without the use of heat or any other 

 change in the system from cycle to cycle, which would consti- 

 tute a perpetual motion machine — a reductio ad absurdum. 

 There is no perfectly reversible engine, but one can be approxi- 

 mated and for the purposes of reasoning one may be postulated. 



We assume that heat has to do with motion of the particles of 

 a body. We have little doubt that matter consists of very small 

 discontinuous particles and there is no reason they should not 

 move. In regard to molecular motion forces are conservative; 

 there are no frictional losses. 



Lecture IV. Continuation of discussion of evidence of fric- 

 tionless character of molecular motion. Count Rumford 

 thought heat not a substance. Joule determined the mechan- 

 ical equivalent of heat; J = 772 ft. pds. W = JQ. We may 

 as well measure Q directly in mechanical units as Q = W. 

 Carnot failed to estabhsh the law Q" = Q' + W, namely, that 

 the difference between the heat received and the heat given up 

 was (proportional to) the work done. Joule seems not to have 

 been entirely clear about the conversion of heat into work. 

 Clausius was the first to set these matters straight. 



Lecture V. Discussion of meaning of first and second laws, 

 and of various ways of stating them, by Tait, Clausius and 

 Kelvin, illustrating each from considerations of the Carnot 

 cycle. If Q" be the heat taken in at one temperature and Q' 

 that given out at the other and W the work done; and if q", q', 

 w be the similar quantities for another engine working between 

 the same temperatures the quantities Q" , Q', W must be pro- 

 portional to q", q', w. For we could by multiphcation (engines 

 in parallel) make Q' = mq'. Now reversing one of the engines 

 (or the set in parallel) the net heat taken or given to the cold 

 reservoir would be nil and if the work were not also nil we should 

 be obtaining work from heat at the single temperature of the hot 

 reservoir which is contrary to Kelvin's statement of the second 

 law. Hence W = imo and since by the first law Q" — Q' = W 



