As an eager scientist on a hot summer day you wish to determine how much ice to
ID: 714950 • Letter: A
Question
As an eager scientist on a hot summer day you wish to determine how much ice to buy to add to your cooler which is filled with 3.0 x 10' cans of soda which are warm at 82.9 F. Each can has a mass of 411 g and ideally you want the temperature of the drinks to be 38.7°F. If there is no heat lost by the cooler and ignoring any heat lost to the soda containers, how much ice needs to be added to your cooler? (Assume e temperature of the ice is 32.0°F.) Number lbs tice Previous Check Answer NextExit Hint This problem is an example of calorimetry. The cooler acts like a calorimeter since there is not heat loss through it. Therefore, the amount of heat transferred to melt the ice and raise the temperature of the melted ice will be equal to the heat to cool the soda cans. The latent heat of fusion for ice is 80.0 cal/g, the specific heat of water is 1.00 cal/lg. "C and the specific heat of the soda is 0.900 calig "C and there are 454 grams per pound.Explanation / Answer
Let us convert temperature into centigrade
We know that
(°F - 32) x .5556 = 0C
82.9 0F = 28.280C
38.7 0F = 3.720C
32 0F = 00C
The heat loss = heat gained
heat gained by ice = heat used in convesion to liquid water + heat to raise the temperature from 00C to 3.720C
Heat gained by ice = mass of ice X heat of fusion + mass of ice X specific heat X change in temperature
Heat loss by cans = mass X specific heat X change in temperature = 30 X 411 X 0.9 X (28.28- 3.72)
Heat loss by cans = 272542.32 Cal
mass of ice X 80 + mass of ice X 1 X (3.72-0) = 272542.32 Cal
Mass of ice ( 80 + 3.72) = 272542.32 Cal
mass of ice = 3255.40 grams
the grams will be converted to lbs as
1 gram = 0.0022 lbs
3255.40 grams = 7.18lbs
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