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OldSpice OldSpice
wrote...
Posts: 6435
7 years ago
Calculate the heat absorbed when 10.0 g of ice at 0 °C melts to water at the same temperature. The specific heat of water is 1.00 cal/(g x °C); the heat of fusion is 80.0cal/g; and the heat of vaporization is 540.0 cal/g.
A) 8.0 cal
B) 10.0 cal
C) 54.0 cal
D) 8.00 x 102 cal
E) 5.40 x 103 cal
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Answer accepted by topic starter
ElectricElectric
wrote...
Top Poster
Posts: 6431
7 years ago Edited: A month ago, bio_man
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D

mass if ice = 10.0 g

temperature = 0 oC

heat of fusion = 80.0 cal / g

heat absorbed = 10 x 80

                       = 800 cal

heat absorbed = 8.00 x 10^2 Cal
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7 years ago
Perfect!

..Thank you so much for your support
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7 years ago
My pleasure
.silent.boss
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A month ago
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