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Specific heat / latent heat

  • Thread starter cugirl
  • Start date
  • #1
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Homework Statement



An insulated cup contains 1kg of water initially at 20 C. 0.50 kg of ice, initially at
0 C is added to the cup of water. The water and ice are allowed to come to thermal equilibrium. The specific heat of ice is 2000 J/kg oC, the specific heat of water 4186 J/kg oC, the latent heat of fusion of water is 33.5x104J/kg. What is the final temperature of the water?

(A) 0 C
(B) 1.2 C
(C) 4.6 C
(D) 9.2 C
(E) 15.2 C



Homework Equations


Q_gain(ice) = Q_lost(water)
mc(Ti-Tf) = mLf + mc(Tf-0)


The Attempt at a Solution


Q_gain(ice) = Q_lost(water)
mc(Ti-Tf) = mLf + mc(Tf-0)
(1.0)(4186)(20 - Tf) = (.5)( 33.5x104 ) + (.5)(4186)(Tf)
83,720 – 4186Tf = 167,500 + 2093 Tf
-83780= 6279 Tf
-13.3 = Tf
 

Answers and Replies

  • #2
mgb_phys
Science Advisor
Homework Helper
7,774
12
Correct method
But what is the total energy needed to melt the ice?
How much energy can you get from the initial water before it cools to 0c?
 
Last edited:

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