Calculate the heat of fusion of ice from the following data for ice at 0- C added to water- Mass of calorimeter -60 gm- mass of calorimeter - water -460 gm- mass of calorimeter - water - ice -618 gm- initial temperature of water -38- C- final temperature of the mixture -5- C- The specific heat of calorimeter -0-10 cal - g - C- Assume that the calorimeter was also at 0- C initially-A- 78 cal - gmB- 78 cal - kgC- 86 cal - kgD- 86 cal - gm

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Byju's Answer

Standard XII

Physics

Latent Heat

Calculate the...

Question

Calculate the heat of fusion of ice from the following data for ice at $0^{\circ} C$ added to water. Mass of calorimeter = 60gm, mass of calorimeter + water = 460 gm, mass of calorimeter + water + ice = 618 gm, initial temperature of water = $38^{\circ} C$ , final temperature of the mixture = $5^{\circ} C$ . The specific heat of calorimeter = $0.10 c a l / g /^{\circ} C$ . Assume that the calorimeter was also at $0^{\circ} C$ initially.

86 cal/kg

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86 cal/gm

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78 cal/kg

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78 cal/gm

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Solution

The correct option is D
78 cal/gm

Mass of water = 460 - 60 = 400gm

Mass of ice = 618 - 460 = 158 gm

Heat lost by water = Heat gained by ice to melt + Heat gained by (water from molten ice + calorimeter) to reach $5^{\circ} C$

$\Rightarrow$ $400 \times 1 \times (38 - 5)$ = $158 \times L + 158 \times 1 \times 5 + 60 \times 0.1 \times 5$ (where $L$ is the latent heat of fusion of ice)

$\Rightarrow$ $L$ = 78.35 cal/gm

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