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

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 cans is 12.5 g of Aluminum. If the temperature of ice is ? 8 Degree C, how many 10- pound bags of ice are needed to chill the cans and their contents to an ce cold 0 degree C ? Assume heat is only transferred between the and not lost to the environment that the ice melts, and that the liquid portion of the beverage has the same pacific heat as liquid water

Explanation / Answer

Let the number of ice bags be y

we know that

1 pound = 453.592 g

so

10 pound = 4535.92 g

so

one ice bag weighs 4535.92 g

we also know that

specific heat of aluminium = 0.9

specific heat of water = 4.184

specific heat of ice = 2.01

we know that

heat lost by hot body = heat gained by cold body

so

heat lost by aluminium cans = heat gained by ice bags

aslo

heat = mass x specific heat x temp change

so

consider

heat lost by aluminium cans = heat gained by ice bags

mal x sal x dTal + ml x sl x dTl = mi x si x dTi

72 ( 12.5 x 0.9 x 25 + 355 x 4.184 x 25 ) = y x 4535.92 x 2.01 x 8

solving we get

y= 36.93

so

36.93 bags of ice is required = 37 bags

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