How much heat (in kj) is needed to convert 866 g of ice at -10C to steam at 126

Question

How much heat (in kj) is needed to convert 866 g of ice at -10C to steam at 126 C? (The specific Heat of Ice, water, and steam are 2.03 j/g C,4.184 j/g C, and 1.99 j/g C respectively.) The heat of fusion of water is 6.01 kj/mol, the heatvaporization is 40.79 kj/mol. How much heat (in kj) is needed to convert 866 g of ice at -10C to steam at 126 C? (The specific Heat of Ice, water, and steam are 2.03 j/g C,4.184 j/g C, and 1.99 j/g C respectively.) The heat of fusion of water is 6.01 kj/mol, the heatvaporization is 40.79 kj/mol.

Explanation / Answer

Bring ice to 0 C q = mass*cp*delT q = 866 g *(2.03 J/g/C) *(0 - (-10))C q = 17579.8 J Melt ice q = delH*mass q = 6.01 kJ/mol *866 g*(1000 J/1kJ)*( 1mol /18g) q = 289147.778 J Bring water to 100 C q = 866 g *4.184 J/g-C *(100 - 0) C q = 362334.4 J Vaporize water q = 866 g*(1 mol/18 g) *(40.79kJ/mol )*(1000 J/1kJ) q = 1962452.22 J bring steam to 126 C q = 866 g *(1.99 J/g-C)*(126 - 100 )C q = 448378.84 J Sum of q = 17579.8 + 289147.778 + 362334.4 + 1962452.22 +448378.84 Q = 3079893.04 J ~ 3.08 e3 kJ

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