CHEMICAL .AND PHYSICAL NOTES. 125 
is exactly 10° above that of the air, and it is found that in the Ist 
second it falls to 9°°9, then at the end of the 1st second the heat 
remaining is only +o of the initial amount which we may represent 
by unity, and in the interval it has lost +1, of the amount that it 
had at the beginning of the interval. If the above principle holds 
good, it must lose in the 2nd second +35 of the amount which it 
has at the beginning of that second, and at the end of the 2nd 
second the heat remaining will be 7% %5 of the amount which was 
present at the beginning of that second. But the amount present 
at the beginning of the 2nd second was ,%% of the amount present 
at the beginning of the 1st second, which is represented by unity. 
Therefore, referred to the amount present at the zero of reckoning as 
unity, the amount remaining at the end of the 2nd second should 
be (4285). Similarly, during the 3rd second the thermometer loses 
aha of the heat which it had at the beginning of that second, and the 
pines remaining at the end of the 3rd second will obviously be 
Pes of (4%95)°, or (4225)%. Similarly, at the end of the 4th, 5th, 6th, or 
ath second, the heat remaining will be (4°5’5)*, (q'95)®, Gaea)®. - G25) 
of the original amount present at the zero of reckoning represented 
as unity. It is obvious that if the law holds good, by giving the 
suitable value to the index » we can at once calculate the proportion 
of heat which will remain after any number of seconds of cooling; 
and it is also obvious that so long as the temperature of the 
medium remains constant, the thermometer can never exactly reach 
that temperature, although the difference of the temperatures may 
be made as small as we like by making the duration of cooling 
sufficiently long. 
It has been said that by giving 2 the suitable value, we can at 
once find the heat remaining after the lapse of any time. But the 
computation of high powers of numbers by ordinary arithmetic is 
very laborious. If, instead of simple arithmetic, we use logarithmic 
arithmetic, the computation of a high power is as easy and expeditious 
as the computation of a low one. In the case which we have imagined 
the constant fraction is +%%,. Its logarithm is log. 99 — log. 100, that 
is 1:9956352 — 2 = 0°9956352 — 1, which is usually written 
1+9956352. 
This is the logarithm of the first power of 7895, which is the heat 
remaining at the end of the 1st second of cooling. If we multiply 
.1°9956352 by 2 we have the logarithm of the square of ,°;, and if we 
multiply it by 3,4... we have the logarithms of the 3rd, 4th... 
nth power of +#%,, that is, of the heat, remaining at the end of the 3rd, 4th 
. .nth second. These logarithms differ from each other by the same 
