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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.)