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Conservation of Heat Energy

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


A 25 g glass tumbler contains 350 mL of water at 24°C. If four 20 g ice cubes each at a temperature of -3°C are dropped into the tumbler, what is the final temperature of the drink? Neglect thermal conduction between the tumbler and the room.

Homework Equations


Q = mcT
Q = mLf (energy required to melt a substance of mass m)

The Attempt at a Solution


Here are all the known variables:
massw = .35kg
c (specific heat for water) = 4186 J/kg * C
Twater,initial = 24 C
massice = 4*20g = .08kg
Lf = 3.33*105 J/kg
Tice,initial = -3 C
Tf = ?

First I can use Q = mcT to see if all the ice will get melted by the water
Q = (.35kg)*(4186 J/kgC)*(24C) = 35162.4 J
m = Q/Lf = (35162.4 J) / (3.33*105 J/kg) = .105592
Since the total mass of the ice is .08, then all the ice will get melted.

Now we can apply conservation of energy to find the final temperature.
massw*c*(Twater,initial - Tf) = massice*Lf + massice*c*(Tf-Tice,initial)

Plugging in all my variables and solving I get 4.18 C which is incorrect.
My math could be wrong but I don't believe it is.
Could anyone provide any insight on what I am doing wrong?
 
Last edited:

Answers and Replies

  • #2
Doc Al
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The initial ice goes to the final water in three steps: Increase the temperature of the ice; melt the ice; increase the temperature of the resulting water.
 
  • #3
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I'm still confused. Here are the three steps you mentioned.

(1) Increase the temperature of the ice: Q = mice*cice*(Tf-Tice,initial)
(2) Melt the ice: Q = mice*Lfusion,water
(3) Increase temperature of water: Q = mwater,initial*cwater*(Twater,initial*Tf)

And then the equation would be: (1) + (2) = (3)

Is this the correct setup?
 
  • #4
Doc Al
Mentor
44,904
1,169
I'm still confused. Here are the three steps you mentioned.

(1) Increase the temperature of the ice: Q = mice*cice*(Tf-Tice,initial)
(2) Melt the ice: Q = mice*Lfusion,water
(3) Increase temperature of water: Q = mwater,initial*cwater*(Twater,initial*Tf)

And then the equation would be: (1) + (2) = (3)

Is this the correct setup?
Not exactly. The ice is warmed up to the final temperature and the 350 ml of water is cooled down to the final temperature. The energy added to the ice equals the energy lost by the water.

Just thinking about the warming of the ice, here are the three steps:
(1) Raise the temperature of the ice from its initial temperature to the melting point.
(2) Melt the ice.
(3) Raise the temperature of the melted ice (water, of course) from the melting point to the final temperature.

Give it another shot.
 
  • #5
15
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I figured it out. Thank you!
 

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