﻿Rate of Evolution of Heat by Orangite. 189 



to the valve having become stack up, but on clearing out the 

 valve air could plainly be heard rushing into the vessel, so 

 that any leakage during the thirteen months for which the 

 vessel was closed must have been small if not actually zero. 



Sources of Error — Possible Effect of Air. 



The internal volume of the copper vessel is about 220 cc. 

 Of this at least 90 cc, must be occupied by the orangite, 

 leaving an air-space of 130 cc. Let as assume as an extreme 

 case that the valve leaked to such an extent that the air- 

 pressure inside rose to that of the atmosphere (which it did 

 not do). This would involve the entry of 130 x 4'835 x 10 -6 

 = 6*3 x 10" 4 gram of water-vapour. Even if each gram of 

 water yielded 840 calories, as much as it would do in com- 

 bining with CaO, the heat so generated would only amount 

 to 0'53 calorie. This quantity is entirely insignificant, as the 

 results obtained indicate a generation of at least 900 calories 

 during the time that the vessel was closed. 



The oxygen entering the calorimeter would amount to 

 39 milligrams, a quantity which, even at the rate of 2000 

 calories per gram, would not seriously affect the result. 



Uniformity of Temperature of Copper Vessel. 



The heat generated by the heater must chiefly reach the 

 walls of the vessel by conduction along the centre tube, which 

 is about 0*7 cm. mean diameter, and is attached to the outer 

 walls above and below. Let us assume that the heater is 

 generating O'l calorie per hour, which is about the value 

 generated on two occasions. Suppose half of this is set free 

 at the upper end of the centre tube round a circle 0"7 cm. in 

 diameter, and let us make the extreme assumption that heat 

 only escapes from the vessel round the rim, which is about 

 7 cm. in diameter. Then for the radial flow of q calories 

 per sec. across a disk of thickness t and conductivity K, we 

 find that 



6 = JL 



V -yy loo- 



2tt[U ° e >V 



where 6 is the difference of temperature between two points 

 distant i\ and r 2 from the centre. In this case 



y= 0'14? 3600, / = 0\15, K=l nearly, and r 2 fr{**lQ t 



so 0=6;8xlQ- 5O C. 



This is onlv about 0-15 scale-division, so it is evident thai 



