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Scholar
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Respect (95%):
Date Posted: 7/24/2008 5:03:10 PM  Status: Closed
calorimetry
Course Textbook Chapter Problem
N/A N/A N/A N/A
Question Details:
A cube of ice is taken from the freezer at -8.5°C and placed in a 75-g aluminum calorimeter filled with 3.1E2 g of water at room temperature of 22.4°C. The final situation is observed to be all water at 19.0°C. What was the mass of the ice cube?
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Answers:

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Oracle
Karma Points: 9,011
Date Posted: 7/24/2008 5:32:24 PM  Status: Live
Asker's Rating: Lifesaver   
Response:
 
   At thermal equilibrium
 
   Heat gained by Water + calorimeter   =   heat lost by ice
 
   mw * cw * Δt   +   ma * ca * Δt   =   mi * ci * Δti   +  mi * cw * Δtw   +  mi * L
 
   Substcript w refers to water, a to aluminum and i to ice.
 
   m   =   mass,   c   =   specific heat,   Δt   =   change in temp of water in caloriemeter
 
   Δti   =   change in temp of ice,   Δtw   =   change in temp. of molten ice
 
(3.1 * 102 * 4.2  + 75 * 0.902) * (22.4 - 19)  =  mi * { 2.1 * (0 - (-8.5) + 4.2 * (19 - 0) + 334)
 
  4656.81   =   mi * 431.65
 
   Mass of ice   mi   =   4656.81 / 431.65   =   10.79   g



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