Question
Asked Dec 1, 2019
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How much heat is required to convert 11.0 g of ice at -14.0 ∘C to steam at 100.0∘C? (Express th anser in joules, calories, and British thermal units)

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Expert Answer

Step 1

Given data:

Mass of the ice, m = 11.0 g = 0.011 kg

Specific heat of the ice, C = 2093 J/kg oC

Specific heat of water, Cw = 4186 J/kg oC

Latent heat of fusion, L = 334 J/g

Latent heat of vaporization, Lvap. = 2260 J/g

Step 2

Heat required to convert ice from – 14.0oC to the 0oC:

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О 3 тСАТ - 0.011 kgx2093 J/Kg'C x(0°с-(-14° с)) =322.322 J

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Step 3

Heat required to convert ice from 0oC to...

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Q, =mL =11 g x 334 J/g 3674 J

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