26 



For example, suppose we wished to find the number of pounds 

 of ice necessary to cool 500 pounds of milk from 80 F. to 40 F. 

 Assuming the specific heat of the milk to be .93, the number of heat 

 units to be extracted from the milk would be 500 X .93 X 40 (80 40 

 = 40) = 18,600. Since the milk is to be cooled to a point 8 F. above 

 the melting point of ice (40 32 = 8), the available refrigeration of a 

 pound of ice would be increased to 152 (144 + 8 = 152). The num- 

 ber of pounds of ice necessary would therefore be 1 8 ,600 -f- 152 = 122.3 

 pounds. 



The theoretical amount of ice will usually be insufficient to cool 

 milk to the desired temperature. This is because of outside influences 

 such as temperature of the air, temperature of the water with which 

 the ice is mixed, and temperature of the apparatus through which the 

 water is run. 



EXPLANATION OF EXERCISE VIII 



Lower temperatures may be obtained with a mixture of salt and 

 ice than with ice alone. This is because of the fact that when two 

 solids unite to form a liquid, they absorb heat. Within certain limits 

 the larger the percentage of salt used, the lower the temperature that 

 can be obtained. After a certain amount of salt has been added the 

 solution becomes saturated and the further addition of salt has no 

 effect in lowering the temperature. 



The following table from B. A. I. Bulletin No. 98 gives the approxi- 

 mate temperatures which may be obtained with different percentages 

 of salt and ice. 



Percentage of salt in Temperature of % of salt in Temperature of salt 



mixture mixture mixture m mixture 



op op 



o 32 15 ii 



5 27 20 1.5 



10 20 25 -IO 



