Calculate the heat of fusion of ice from the following data for ice at $$0$$ºC added to water. Mass of calorimeter $$=$$ $$60$$ g, mass of calorimeter $$+$$ water $$=$$ $$460$$ g, mass of calorimeter $$+$$ water $$+$$ ice $$=$$ $$618$$ g, initial temperature of water $$=$$ $$38$$°C, temperature of the mixture $$=$$ $$5$$°C. The specific heat of calorimeter $$=$$ $$0.10$$ ca $$lg-1$$ °$$C-1$$.

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Infinity Learn · Accepted Answer

Mass of water $$=$$ $$460$$ $$-60$$ $$=$$ $$400$$ gMass of ice $$=$$ $$618$$ $$-$$ $$460$$ $$=158$$ gHeat lost by water $$=$$ Heat gained by ice to melt $$+Heat$$ gained by (water$$ $$+$$ $$calorimeter) to reach $$5$$ºCÞ $$400$$×$$1$$×(38-5) $$=158$$×$$L+158$$×$$1$$×$$5$$ $$+$$ $$60$$×$$0.1$$×$$5(where$$ L is the latent heat of fusion of $$ice)$$⇒ L $$=$$ $$78.35$$ calg–$$1$$

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