FIND THE LATENT HEAT OF FUSION OF WATER ? PLEASE SHOW THE WORK Given value, Room

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FIND THE LATENT HEAT OF FUSION OF WATER ?   PLEASE SHOW THE WORK
Given value, Room temperature = 25.4 degree C mass of empty calorimeter cup (mcal) = 7.9 grmas Initial temperature of ice (Ti) = -16.9 degree C Initial temperature of water Twi= 25.4 degree C Final temperature in calorimeter(Tf) = 19.5 degree C mcal+mw= 259.8 grmas mcal +mw + mice = 274.9 grams mw= 251.9 grams mice= 15.1 grams ACCEPTED VALUE FOR LATENT HEAT OF FUSION OF WATER IS 334 KJ/KG. SO FIND THE PERCENT ERROR. PLEASE SHOW ALL THE WORK    Thanks .

Explanation / Answer

we know that, the specific heat of ice = 0.500 cal/ g'C the specific heat of water 1.00 cal/g'c the specific heat of calorimetry = 0.215 in given formula, 0.5*15.1(16.9) + 15.1 Lfw + 15.1 (9.5) = 1.00*251.9(5.9) + 0.125*7.9(5.9) 15.1 Lfm = 1120.796 Lfw = 74.22 cal/g'C therefore Lfw = 310 kJ/kg 'C the error is 24 kj/kg.

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