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 $$lg-1$$ °$$C-1$$.

Question

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–$$1$$

Calorimetry

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

Calorimetry

Remember concepts with our Masterclasses.

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 5ºCÞ 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

Ready to Test Your Skills?

free study material