170 
IOWA ACADEMY OF SCIENCE 
If r comes out minus we consider that no HNOg is formed and proceed to 
determine the HNO 3 used for N 2 O 5 and NO 2 . To determine this we calculate 
the amount of HNO3 used for all the zinc and subtract that from the HNO3 used 
and that leaves the HNO3 decomposed into Hg and N2O5. These results are 
tabulated in data No. 4. 
As we have the ratio of the different reductions of the HNO3 we can deter- 
mine n and then the calculation of E' is very simple by subtracting in equa- 
tion (9). This is done for the five intervals of time and tabulated in first part 
of data No. 7. 
In calculating the E. M. F. (E') of the cell at different times we have not 
taken into consideration the temperature coefficient (deldt) which must enter 
into the equation for the time calculated E. M. F. We must now direct our 
attention to the temperature coefficient (de|dt). 
In doing this we will let H' equal to the amount of heat supplied (in ergs) 
at the temperature T' and H" the heat subtracted at the temperature T". Then 
(H'— H" H' = (V —T") + T' 
H' — H" is the quantity of heat used and is equal to O" (E' — E") ; where 
O" equals the units of electricity passed at T' and T" and E' and E" the E. M. 
F. at T' and T". 
Therefore O" (E'— E") - H' = (T'— T") 4 - T' 
thus H' = O" T' (E'— E") -4 (T'— T"). 
(E' — E") 4 - (T' — T") is the rate of change of the E. M. F. of cell with tempera- 
ture and we will designate it by de 4 -dt. By letting O" equal to unity the equa- 
tion reduces to 
H = T (de-4dt) (14) 
Which is the quantity of heat converted into electrical energy during the 
passage of a unit quantity of electricity. It is to be noted that deldt carries 
its sign. That is it is plus if it increases with temperature and minus if it 
decreases with temperature. 
If H'" is total amount of heat produced by chemical action when one unit 
quantity of electricity passes through the cell. Then 
H'" -t- p = E 
When H'" and p are in ergs. 
Consequently, H'" 4- de4-dt = E (15) 
And E is the time calculated E. M. F. of cell temperature T on the absolute 
scale. 
Now we will proceed to find de|dt, which is necessary in calculating the time 
E. M. F. We will represent the current by c and the P. D. at the terminals of 
cell on closed circuit with external resistance r by e. Then the heat developed 
in the external circuit in the time t is equal to aect calories, when (a) equals 
.2387, which is the value of one joule in calories. The heat developed within 
the cell itself in the time t, by the passage of the current (c) is ar"e^t where v" 
is the internal resistance of the cell. Therefore (ar"c^t + aect) is the amount 
of heat developed by the passage of ct units of electricity. 
E = s e 4- cr" 
multiply by act 
acEt = acet 4- ac^rt (16) 
It will be noticed that each side of this equation represents the energy trans- 
ferred into heat. 
