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navy_bison
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Homework Statement
How much heat is necessary to change 395 g of ice at -10°C to water at 20°C?
Homework Equations
mLf + mc(delta)T
The Attempt at a Solution
(.395kg)(80kcal/kg)+(.395kg)(1.00kcal/kg x 1 deg C)(20--10)
(31.6kcal)+(.395kcal degC)(30)
31.6kcal + 11.85 kcal=43.45 kcal
I got the answer wrong obviously, the correct answer should be 41.5 kcal. Please help me, my teacher isn't so great and an explanation as to where I went astray would be appreciated. Thanks
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