BY JAMES TOLSON, ESQ. 63 
we have 1'045 X 180 = 188'1 units as the quantity of heat radiating 
per square foot per hour from the surface of cast iron, if it be kept 
at a constant temperature.—Box, op. cit. 
In the case where the temperature of the absorbent is 
below zero centigrade, the formula becomes, 
R’=124-72 x 10077. x (1067721) 
Ree 
pee ae 
=124°72 X 10077" X(1'0077 7 I). 
Ai 
In which -t=the number of degrees below zero centigrade 
of the absorbent. As an example, take a body at a tem- 
perature of 95° F. or 35° C., hanging in a freezing room, 
the walls of which are at-—13° F. or—25° C., the formula 
then becomes— 
124°72 X 10077” X (1'0077 7) 
60 
1 
=124°72 X 1214 X (1'584—1). 
60 
= 1°'004 = ratio of loss of heat. 
Taking meat, at a temperature of 95°, as an example, having 
a radiating power of, say, 1°25 (by Table No. 1), the loss 
per square foot, for a difference of temperature between the 
walls and the meat of 108° F. or 60° C., would be at the 
rate of 1°25 X 1°004 X 108 = 135'5 units per square foot 
per hour at the commencement. As, however, the supply 
of heat in the meat is only limited, and it is not kept up, 
the radiation falls off from the first moment, until it finally 
ceases, when the temperature of the meat and the walls are 
the same. 
The following table gives the ratio of the loss of heat by 
radiation, for temperatures from —40° to 104° Fah. of the 
absorbent, which in our case is the wall of the freezing 
chamber, and with a difference of from 9° to 135° Fah. in 
the temperature of the radiant and absorbent—the meat 
and the walls. It is necessary to use a table of logarithms 
for raising the numbers to the powers given in the above 
calculations. 
