446 PROFESSOR A. SCHUSTER AND MR. W. GANNON ON A 
Also ¢$(t) = pt between ¢ = 0 andt=T, and ¢(t) = pT itt>T 
The solution of the differential equation can be put into the form 


7 Qn=n 7] ’ + » 2 2 = 0 
u=$(t)" oa = > sin et + SES 
. IT Dh aes 
3) GS — ae | Bae MINI GS ON GDN: 
T n=1 1 C T n=1 ¢ 0 
where £ is written for az/e. 
The first term is often omitted as it vanishes for all finite values of x But as we 
shall have to consider the value of du/ox for « = 0, it is retamed here. To perform 
the integration we must distinguish two cases, according as ¢ is smaller or greater 
than T. Taking ¢(A) = pd, and performing the integration with respect to A 
we find 
C—2 2p 2 il 
NTL 
y= (1 — e-”*") sin 
C TB? na 8 c 

U = pt 

(1) 
For larger values of t we have to divide the integral into two parts as the value of 
¢ (A) has a discontinuity for ¢ = T. 
The final equation for u then becomes 
(tonto 
Dn fo} 
mee 2p 1 pant et (pn2per oe MUTT 
epi — — F =e" (e*PT — 1) sin —- 
L c TB? na 12 ( ) c 

Writing (du/ox), for the value of du/ox at the plane « = 0, the quantity of heat 
passing out of the calorimeter up to the time T will depend on 
T T 
aa \ 2 dt = = | jer AR 2p i) spl — gory de 
CX /0 -) np? 
0 0 
T eT 5 2 » N= oO 1 
ne Pp oe pe i 30> ae 2pe S e 
2¢ 302 «AB at le an WE 



— RT 
We also obtain ift > T 

t 
Ou T be ne Qn 2 I i ie 
~\) dt 7 (2t =D) ag ee Se (cr) 
\OL /o aC 
3a Ta* y= 1 n* 
0 
The application of these formule is easy when ft is so large that e~*¢~” 
may be neglected. The heat H which has passed through unit surface of the 
calorimeter up to the time ¢ is obtained by multiplying the right-hand side of the 
last equation by k = a’s. 
Neglecting the last term this becomes 
ee 

= (2t a T) + ue 
