Specific latent heat and heat capacity

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SUMMARY

The discussion focuses on a thermodynamics problem involving the melting of ice in lemonade. Given the specific latent heat of fusion of ice at 336,000 J/kg and the specific heat capacity of lemonade at 4,200 J/kg, the calculations show that the total energy loss by the ice cubes is 33,600 J. The heat capacity of the lemonade is calculated to be 1,386 J, leading to a temperature drop of approximately 24.24 degrees Celsius. Consequently, the final temperature of the lemonade after the ice has melted is determined to be 3.80 degrees Celsius.

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Homework Statement



0.1kg of ice was added to a lemonade of mass 330g at an initial temperature of 28 degree celsius and it completely melted. Considering only the lemonade and ice cubes, calculate the final temperature of the lemonade.

Given: specific latent heat of fusion of ice= 336000 J/kg
specific heat capacity of lemonade= 4200 J/kg


Homework Equations





The Attempt at a Solution



total energy loss by ice cubes= 0.1kg * 336,000J
=33,600J

heat capacity of 330g of lemonade=0.33kg * 4200 J/kg
= 1,386J

Total temperature fall by lemonade=33,600J / 1,386J
=24.24242424

Final temperature=Initial temperature - fall in temperature
= 28-24.242424
= 3.80 degree celsius (3 s.f.)
 
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