Anonymous
Anonymous asked in Science & MathematicsChemistry · 5 months ago

Can you help me with these questions?

How many calories are required to melt 75 grams of ice at 0 degrees Celsius? & How many calories are required to evaporate 75 grams of water at 100 degrees Celsius?

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  • 5 months ago
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    334 J/g x 75 g x 1 cal/4.184 J = 5990 cal

    2256 J/g x 75 g x 1 cal/4.184 J = 40400 cal

    you do know that the calorie is an obsolete unit?

    specific heat of water is 4.186 kJ/kgC = 4.186 J/gC

    = 1 calorie/gC = 5.375 J/mol·K = 1 Btu/lb-F

    specific heat of ice is 2.06 kJ/kgC = 2.06 J/gC

    specific heat of steam is 2.1 kJ/kgK = 2.1 J/gK

    heat of fusion of ice is 334 kJ/kg = 334 J/g

    heat of vaporization of water is 2256 kJ/kg = 2256 J/g

  • 5 months ago

    (333.6 J/g) x (75 g) / (4.184 J/cal) = 5980 cal to melt

    (2257 J/g) x (75 g) / (4.184 J/cal) = 40458 cal to evaporate

    [Adjust the number of significant digits as you see fit.]

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