During a picnic on a hot summer day, all the cold drinks disappeared quickly, an

Question

During a picnic on a hot summer day, all the cold drinks disappeared quickly, and the only available drinks were those at the ambient temperature of 80 degree F. In an effort to cool a 12-fluid-oz drink in a can, a person grabbed the can and started shaking it in the iced water of the chest at 32 degree F. Using the properties of water for the drink, and neglecting heat transfer to the soda can and to the hand, determine the mass of ice that will melt by the time the canned drink cools to 50 degree F. Please give the answer in kg.

Explanation / Answer

T cools to = 50°F

T ice = 32°F

Tamb = 80°C

so...

Assume:

Qlost by can = m*C*(Tf-Ti)

12 oz drink = 12 oz water ( assume)

12 oz = 355 mL = 355 g (1 g per mL density of water)

dT = 80-50 F = 30 F = 16.67 °C

Qlost = 355*4.184*(16.67) = 24760.2844 J were lost

Those then were gained byt ice

24760.2844 J gained

assume only melting of ice:

Q = m*LHmelting

LHmelting = 334 J/g

so

m = Q/LHmelt = 24760.2844 /334 = 74.132 g of ice will melt

74.132*10^-3 kg = 0.0074 kg of ice melt

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