You are throwing a party and you need to chill your beverages, which consists of

Question

You are throwing a party and you need to chill your beverages, which consists of 72 aluminum cans each containing 355mL of liquid. Assume an aluminum can is 12.5 g Aluminum. If the temperature of the beverage is 25 degrees celsius initially, and the temperature of the ice is -8 degrees celsius, how many 10 pound bags of ice are needed to chill the cans and their contents to an ice cold 0 degree celsius?

(Assume the heat is only transferred between the ice and cans and not lost to the environment, that the ice melts and that the liquid portion of the beverage has the same specific heat as liquid water)

*Give an explanation of each step if possible

Explanation / Answer

First we must first understand that each can has 12.5 g, also also assume that the drink has 355ml, include missing data as latent heat of fusion of ice and specific heat of ice and drink, also drink density tranformar beverage content gr. so give you a standard water values:
density = 1 mL / g ... 355ml / 355g * 355g ... 72 = 25560g
Al gr = 12.5g * 72 = 900 ... 25560g + 900g = 26460g = mass of the drink.

Drink specific heat = 1 cal / (g.

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