You are throwing a party and you need to chill your beverages, which consist of
ID: 813644 • Letter: Y
Question
You are throwing a party and you need to chill your beverages, which consist of 72 aluminum cans each containing 355 mL of liquid. Assume an aluminum can is 12.5 g of aluminum. If the temperature of the beverages is 25 degrees celcius initially, and the temperature of the ice is -8 degrees celcius, how many 10 pound bags of ice are needed to chill the cans and their contents to an ice cold 0 degrees celcius. Assume heat is only transferred between the ice and cans and not lost to enviroment, that the ice melts, and the liquid portion of the beverage has the same specific heat as liquid water.
Explanation / Answer
The Al cans are chilled to zero degrees by cooling in ice.
Thus the heat absorbed by cans/contents = heat given by ice.
Heat absorbed by cans = heat by Al + heat by liquid
= 72 x 12.5 x 25 x 0.9 J + 72 x 355 x 25 x 4.184 J
=20250 + 2673576 = 2693826 J
Heat supplied from ice = mass x 4.184 x 8
Equalizing both and solving for mass of water we get mass of water =80479.9 g =177.42 pounds
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.