Question
Asked Aug 26, 2019
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When ice melts, it absorbs 0.33 kJ per gram.

How much ice is required to cool a 13.0 oz drink from 80 ∘F to 40 ∘F, if the heat capacity of the drink is 4.18 J/g∘C? (Assume that the heat transfer is 100 % efficient.)        
Express your answer using two significant figures.
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Expert Answer

Step 1

First, the grams of drink is determined and then the given temperature in Fahrenheit is converted into Celsius as follows,

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=1307x 28.3 g 1oz 367.9 g 80 F 26.7'C 40°F 4.44 C

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Step 2

Finally, the amount of ice require...

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Q m c AT 367.9 g x 4.18(26.7°C-4.44°C) g°C J x(22.26°C) 367.9 g x 4.18. g°C 34231.9 J - 34.2319 kJ 1 g absorbs 0.33kJ heat 1g = 34.2319 kJ0.33 kJ 103.7 g 100 g

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