BY JAMES TOLSON, ESQ. 97 
Calculating the amount of heat to be abstracted from a 
quantity of meat put into such a room, as illustrated by the 
last shipment sent from the Queensport Works— 
Daily average quantity of meat put into the freez-)_ ibs. 
ing rooms—4z2 quarters, weighing 180 lbs. } 7,560 
Average temperature, 85° 
Specific heat eT Hek 6] 
Heat required to be abstracted— 
Lista eo bring down. the temperature from 85° Units. 
to 32° = 53°, 7,560 X 53 x °533 a uSsoe2 
2nd. To freeze the water contained in 7, sc 
Ibsis-2, 728 ibs.-< 14:2°6 oe ©} 388, ae 
3rd. To reduce 7,560 lbs. from 32° to zero, 
85,397 
7500 32 X 353) (353) = specific 
heat of frozen meat... . 
686,974 
Or g1 units per pound of meat. 
The total quantity of air circulating through the rooms per day 
of 24 hours (23 hours work), 
37,500 X 23 = 862,500 cubic feet. 
(Taken at a temperature of 50° Fah. after passing over the drying 
pipes.) 
Quantity of heat to be removed per cubic foot te 
of air, 686,974 + 862,500 0°7966 
Thermal equivalent of air at ae a column ib. 
g in table 3 : * o'0185 
Rise in temperature in degrees Fah., due to 6 
the absorption of °7966 unit + o185 43°3 
Rise in Cae ae” due to siete through 7c 
the walls . = ar 3 
Rise in temperature due to the abstraction of : 
heat from the meat 43°3 
Total rise etna is 56-3" 
