As an eager scientist on a hot summer day you wish to determine how much ice to
ID: 1057250 • 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 34 cans of soda which are warm at 82.9°F. Each can has a mass of 402 g and ideally you want the temperature of the drinks to be 40.1°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 the temperature of the ice is 32.0°F.) Number 8.49 lbs m ice Previous Check Answer Next E B Hint is 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/g. °C and the specific heat of the soda is 0.900 cal/g. OC and there are 454 grams per poundExplanation / Answer
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 34 cans of soda which are warm at 82.9°F. Each can has a mass of 402 g and ideally you want the temperature of the drinks to be 40.1°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 the temperature of the ice is 32.0°F.)
(82.9 °F - 40.1°F) x (5/9) = 23.78 °C change
Supposing the specific heat of soda is the same as that of water: =q=mst
(4.184 J/g·°C) x (34 x 402 g) x (23.78 °C) = 1359904.767J required
40.1°F = 4.5°C
he latent heat of melting ice is 80 cal/gram, and the specific heat of water is about 1.00 cal/(gram K
Q gain = m(80) + m(1)(4.5) = 84.11111m
Q gain = (m ) (80) + (m ) (1)(4.5) = 84.5m
Q gain = Q lost
m=(1359904.767J J) / (84.5) =16093.54= 16.093 kg of ice
16.093 kg of ice =35.47lb of ice
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