Specific latent heat and heat capacity

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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 celcius 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 celcius (3 s.f.)

 

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