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Change in Temperature Problem

  • Thread starter mopar969
  • Start date
  • #1
201
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Here is the problem:
Take 100g aluminum cup at 25 degrees celcius and 200g of water (h2o) at 25 degrees celcius and add 20 g ice at -10 degrees celcius. Stir.
What is the final state of the system?

Here is what I have done:
I used the equation delta Q= (mcup)(Ccup)(delta T cup)+(mh2o)(Ch20)(delta T h20) + (m ice)(Cice)(delta T ice). And for my final answer I got 23.49 degrees celcius

Is this answer correct and is this all that the question is asking for?
 

Answers and Replies

  • #2
911
18
Hi

You should consider amount of heat energy lost and amount of heat energy gained.
Cup and the water are at higher temperature. and ice is at subzero temperature. So heat energy will flow from cup+water to the ice. This gained energy will be used by the ice , first to raise its temperature from - 10o C to 0o C. At 0o C ,
there will be phase transition. Till all the ice melts, temperature of the ice will not change.
So energy will be used to do phase transition.
 
  • #3
201
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Okay that makes sense but how do I calculate that?
 
  • #4
911
18
Lets first think about the ice. The energy gained from the aluminium cup and the water will be used for heating ice. Lets first find out how much energy is required to completely melt the ice.

Specific heat of ice = 2090 J/kg-C = cice
Latent heat of fusion for ice = [itex] 3.33 \times 10^5 J/kg [/itex] =Lf
mass of ice = 0.02 kg = mice

so amount of energy required to completely melt the ice is

[tex] Q_{ice}= m_{ice}\left[ c_{ice}(\Delta T) +L_f \right] [/tex]

where [itex] \Delta T = 10^{\circ} C = 0^{\circ}C - (-10^{\circ} C) [/itex]

I got Qice = 7078 J.

Now specific heat of aluminium = cAl = 900 J/kg-C
specific heat of water = cw = 4186 J/kg-C
mass of Aluminium = mAl = 0.1 kg
mass of water = mw = 0.2 kg

Now if the aluminium cup + water were to come to the 0 celcius, the amount of energy which will be released is

[tex] Q_2 = m_{Al}c_{Al}(25-0) + m_w c_w (25-0) [/tex]

which is eqaul to 23180 J. But we need only 7078 J to completely melt the ice. So
rest of the energy (23180- 7078 = 16102 J) will be used to increase the temperature
of the melted water. Lets say the equivalent temperature of the whole thing is T. so we will now write the equation where left hand side is the total energy gained by the ice and the right hand side is the total energy lost by Aluminium + water

[tex] 7078 + m_{ice} c_w (T-0) = m_{Al} c_{Al} (25 -T) + m_w c_w (25-T) [/tex]


solving which I got [itex] T = 15.92^{\circ} C [/itex]

see if you can follow this
 
  • #5
201
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Why do you need to add the Latent heat of fusion to calculate the amount of energy to melt the ice?
 
  • #6
gneill
Mentor
20,793
2,773
Why do you need to add the Latent heat of fusion to calculate the amount of energy to melt the ice?
Because the latent heat of fusion is the energy required to turn ice at 0C into water at 0C.
 
  • #7
201
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Okay thanks. But my other problem is that when I try to solve for T I get 2.5. Can anyone please show how the 15 degrees was calculated?
 
  • #8
gneill
Mentor
20,793
2,773
It's pure algebra. How about you show us your step by step attempt, and we'll see if we can spot the error.
 
  • #9
911
18
Use Latex. equations are more readable that way... its easy to learn latex
 

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