282 HEAT. 



taking AB as a straight line, and the area LABML as 110 per gramme 

 of water. We have the values 



BONM = latent heat at 413, or 140 C. 



= 606-5 - -695 x 140 = 508 (Regnault) 



AL = 303; BM = 413. 

 To find ABK 



ABK = IAK-BK = ^AK(BM - AL), 



2 2 



also ABML=!AK(BM+AL)=HO, 



2 



A -ry-n- BM - AL A T.T..T 



whence ABK= - , ABML 



110 

 -jj- x508 



110x110 110 ~a 



Then the efficiency = - ; -- 1- x 508 



716 413 _ 



618 

 = 0-246 about. 



If the cycle were a Carnot cycle, with heat put in only at the higher 

 and taken out only at the lower temperature, the efficiency would be 



This is only a particular cycle, and the efficiency might be brought still 

 nearer the Carnot cycle value by allowing, say, isothermal expansion of 

 the steam at 413, until the pressure was such that, on further adiabatic 

 expansion till the temperature fell to 303, condensation was then just 

 beginning. But such a cycle would be less like that in the actual steam- 

 engine than the one considered. 



The engine we have imagined is a perfect one for the particular 

 cycle, for we have supposed no loss by conduction or friction. Let us 

 now see how much coal would be required per horse-power per hour in 

 such an engine. 



We may estimate that one gramme of coal gives a supply of 8000 

 calories in the source, or 8000 x 4'2 x 10 7 ergs. 



Of this only 0-246 is transformed by our engine, or 1 gramme of coal 

 yields 8-27 x 10 10 ergs as work. 



