Mixing Ice and Water: Final Temp & Composition

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SUMMARY

The final temperature and composition of a mixture formed by adding 300 grams of water at 25°C to 100 grams of ice at 0°C results in 393.75 grams of water and 6.25 grams of ice at 0°C. The heat released by the water, calculated as 7500 calories, is insufficient to completely melt the ice, which requires 8000 calories. Consequently, only 93.75 grams of ice melts, confirming the final state of the mixture.

PREREQUISITES
  • Understanding of specific heat capacity, particularly for water
  • Knowledge of latent heat of fusion, specifically for ice
  • Basic principles of heat transfer in thermodynamics
  • Ability to perform calorimetry calculations
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  • Learn about calorimetry and its applications in phase changes
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Homework Statement



300gm of water at 25°C is added to 100gm of ice at 0°C ,find the final temperature of the mixture .Also find the final composition of mixture .

Homework Equations


[/B]
Specific heat of water = 1 cal/gm/°C
Latent heat of fusion of ice = 80 cal/gm

The Attempt at a Solution


[/B]
Heat released by 300gm water to reach 0°C = 300x1x25 = 7500 cal

Heat required by ice to melt completely
= 100x80= 8000 cal

Since heat supplied is less than heat required , whole ice will not melt .

Amount of ice melted = 7500/80 = 93.75gm

Final composition of mixture = 393.75gm water +6.25gm ice at 0°C

Is my answer correct ?
 
Last edited:
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It is correct.
 
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