79° PRESERVATION OF FOOD BY REFRIGERATION ; 
operation will be at the rate of 1°25 X65 x 1:02=84'15 
units per square foot per hour. 
The loss of heat by contact of cold air will be (by Table 
3), for a difference of 66° F., at the ratio of 1276, and 
the value of A being (by Table 4), for a plane 1 foot high, 
"5945 unit per square foot per hour, we have °5945 Xx 66 x 
1°278=50'145 units as the rate of loss of heat by contact of 
cold air per square foot per hour at the commencement of 
the operation. The total loss, therefore, from radiation 
and contact of cold air per square foot per hour will be at 
the rate of 84°15 +50°145=134'295 units at the commence- 
ment of the operation ; but as the quantity of heat is limited 
to what is contained in the meat, the amount emitted falls 
off from the very commencement of the cooling. 
Loss or Herat sy Conpuction.—“ Let P (fig. 5, plate IL) be 
a plate of building stone 1 foot square, 1 inch thick, having its. 
surface, S, maintained at 60°, and let the walls, W W W, and the 
air in contact with S, be at 60° also, while the surface, S’, is at 59°, 
there will be no loss by radiation and contact of air, because 
both the walls, W W W, and the air are at the same temperature 
as S, but a certain amount of heat will be transmitted from S to S’; 
and for stone 1 inch thick the loss will be 13°7 units per square 
foot per hour for 1° difference. This amount will vary greatly 
with the nature of the material, we will call it C, and its value is 
given in the following table. The amount of heat lost, also, 
varies directly as the difference of the temperature of the two 
surfaces, S and S’, and inversely as the thickness, and hence we 
have the rule— 
C=C xd = 
When C’ = the loss by conduction in units per square foot per 
hour, C = the conducting power of the material, EK the thickness. 
of the plate in inches, and d the difference of temperature between 
Sand S$’. Thus, a stone wall, 20 inches thick, having one surface 
at 70° and the other at 40°, taking the value of C at 13°7, will 
transmit 13°7 x (7O—40) + 20 = 20°55 units per square foot per 
hour.”’—Box, op. cit. 
