COOLING, STORING, AND SHIPPING MILK. 3 
of water at an initial temperature of 37° F. The weight of the can 
containing the milk is taken at 21 pounds and its temperature is 
approximately that of the milk, 85° F. The weights of the water, 
milk, and can are 250, 86, and 21 pounds, respectively, and their 
specific heats are 1, 0.93, and 0.113, respectively. Substituting these 
in the formula above and solving for the final temperature of the 
whole, we have: 
(250X187) + (86 X0.93 X85) + (21011385) _ 16252 
= = 48.9° F. 
(250 X1) + (86 X 0.93) + (210.118) 332. 4 
C= 
In practice, however, some heat is absorbed in the surrounding air 
so that the final temperature of milk and water is higher than that 
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Fig. 1.—Initial and final temperature of milk and water in a concrete-insulated cool- 
ing tank. 
given in the formula. In other words, the final temperature as 
calculated must be divided by the per cent efficiency of the tank. The 
efficiency of a good insulated tank of the size and construction of 
that shown in figure i was found to be about 97 per cent. The final 
temperature of the milk and water will, consequently, be 48.9 divided 
by 0.97 or 50.4° F. 
Ice is usually necessary in order to lower the temperature of water 
to a point at which it will cool milk quickly and efficiently. One 
pound of ice has many times the cooling capacity of an equal weight 
of water at an initial temperature of 37° F. To change the tempera- 
ture of 1 pound of water 1 degree requires the addition or extraction 
of only 1B. t.u. To freeze 1 pound of water, however, requires the 
