Calorimeter Problem: Final Temp of Water & Ice Mixture

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SUMMARY

The final temperature of a mixture containing 300 g of water at 75°C and 20 g of ice at 0°C in a 200 g aluminum calorimeter is calculated to be 66.5°C. The heat lost by the water is determined using the formula Q = mcΔT, resulting in 94,185 J. The heat required to melt the ice is calculated using the latent heat of fusion (Lf = 33.5 x 104 J/kg), yielding 6,700 J. The heat exchange involving the aluminum calorimeter is also accounted for, confirming the final temperature of the mixture as 66.6°C.

PREREQUISITES
  • Understanding of heat transfer principles
  • Knowledge of specific heat capacity (Cal = 900 J/kg°C)
  • Familiarity with latent heat of fusion (Lf = 33.5 x 104 J/kg)
  • Ability to apply the conservation of energy concept
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  • Study the principles of calorimetry and heat exchange
  • Learn about specific heat capacities of various materials
  • Explore the calculations involved in phase changes and latent heat
  • Investigate advanced calorimetry techniques for more complex mixtures
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Students studying thermodynamics, physics educators, and anyone involved in calorimetry experiments or heat transfer 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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