Calorimeter Problem: Final Temp of Water & Ice Mixture

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A calorimeter problem involves a 200 g aluminum calorimeter containing 300 g of water at 75°C, into which 20 g of ice at 0°C is added. The heat lost by the water as it cools to 0°C is calculated to be 94,185 J, while the heat required to melt the ice is 6,700 J. After accounting for the heat exchange, the final temperature is estimated to be around 66.5°C. The calculations also incorporate the heat absorbed by the aluminum, leading to a refined final temperature of approximately 66.6°C. The problem emphasizes the importance of considering all components in thermal equilibrium calculations.
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Homework Statement


A 200.0 g aluminum calorimeter holds 300g of water at 75 C. 20 g of ice at 0 C are added to the water. What is the final temperature of the mixture?


Homework Equations


Use Lf= 33.5 x 104
Cal= 900 J/kg C

The answer is 66.5 C

The Attempt at a Solution


Don't know where to start.
 
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Well, this is what i came to:

Find the heat lost by the water, if it is cooled to 0 C:
(.3)(4186)(75) = 94185

Calculate the heat needed to melt the ice:
(.02)(33.5e4) = 6700Q left:
( 94185 - 6700) = 87485

Final temp = 87485 / ( .32)(4186) = 65.3 C

The answer is 66.5 C. I don't know what to do with the aluminum
 
Woohoo I think I got it:

QAL = (.2)(900)(65.3-75) = 1746

Final temp = (1746 + 87485) / (.32*4186) = 66.6 C
 
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