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How much heat energy, in kilojoules, is required to convert 48.0 g of ice at −18.0...

How much heat energy, in kilojoules, is required to convert 48.0 g of ice at −18.0 ∘C to water at  25.0 ∘C ?

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Answer #1

ice converted to water at 25C0 in three steps

q1 = mCT

       = 48*2.09*(0-*-18)   = 1805.76J

q2    = mH fustion of ice

       = 48*334 = 16032J

q3   = mCT

       = 48*4.18*(25-0) = 5016J

q   = q1 +q2 +q3

      = 1805.76 +16032 +5016 = 22853.76J   = 22.85376KJ

         

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