2 BULLETIN 744, U. S. DEPARTMENT OF AGRICULTURE. 
1° F., it takes only 0.113 B. t. u. to raise 1 pound of iron 1° F. The 
specific heat of a substance is its ability or capacity, compared with 
water, to absorb heat. 
All substances in nature contain more or less heat, but no way has 
been found for abstracting all the heat contained. In practice, how- 
ever, we are not concerned with the total amount of heat a body may 
contain but with the changes that occur in the quantity; that is, how 
much heat is added or subtracted in raising or lowering the body 
through a given range of temperature. This is found by multiply- 
ing the weight of the body in pounds by its specific heat and then 
multiplying the product by the number of degrees change in tempera- 
ture. For instance, if 100 pounds of warm water were cooled 100° F. 
the heat given up would be 100*1X100=10,000 B. t. u. 
The average specific heat of milk which contains 34 per cent but- 
terfat is 0.93; therefore, the amount of heat required to raise 100 
pounds of such milk 50° F. is 1000.93 x50=4,650 B. t. u. If a like 
quantity of milk is cooled the same number of degrees, the B. t. u. 
given up are the same. In other words, the calculations are identi- 
cal whether milk is heated or cooled. 
Air and water are two natural cooling agencies. Cooling by means 
of air is not generally practicable because of its low heat-absorbing 
capacity and because its temperature is usually too high. Water, 
therefore, is the common cooling agency used. If it were possible 
to avoid all losses, 1 pound of milk in being cooled 1° F. would re- 
quire 0.93 of a pound of water which would be raised 1° F. in tem- 
perature. In practice, however, a certain amount of cooling effect is 
always lost through radiation; consequently more water will be re- 
quired, the exact quantity depending upon the efficiency of the cool- 
ing apparatus. This is assuming that the cooler was 100 per cent 
efficient. To determine the final temperature when milk is cooled by 
water and allowed to remain until both are practically the same tem- 
perature the following formula may be used: 
_— (WXSXT) + (WSsTs) + (W28:T2) 
WS+W:S:+ W282 
t=—Final temperature of whole. 
W=Weight of water. 
S=Specific heat of water. 
T=Initial temperature of water. 
W:i=Weight of milk. 
S:=Specific heat of milk. 
T:—Initial temperature of milk. 
W:.= Weight of can. 
S.2=Specifie heat of can. 
T.=—Final temperature of can. 
For example, figure 1 shows a 10-gallon can of milk at an initial 
temperature of 85° F. being cooled in a tank containing 30 gallons 
