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Final Temperature After Phase Change

  • Thread starter Newton86
  • Start date
  • #1
59
0

Homework Statement



find the final temperature of A 150g (0,15kg) cube of ice at 0.0 degrees Celsius is added to 300g(0.3 kg) of water at 50.0 degrees Celsius.

Specific heat capacity of ice: 2.09 x 10^3 J/(kg x degrees Celsius)

Specific heat capacity of water: 4.186 x 10^3 J/(kg x degrees Celsius)



Homework Equations




Q = m x L
energy transferred as heat during a phase change = mass x latent heat

Q = m Cp deltaT
energy transferred as heat = mass x specific heat capacity x change in temperature



The Attempt at a Solution



I wouldnt had problem to solve if it wasnt for the phase change :cry:

For the water q=mcdeltaT
(0,3kg)*(4190j/kg*k)(T-50C)
If it was no phase its the same for the other liquid and then just set them equal to = 0 and find the T:smile: but no with the Phase it will be a bit different but how ?

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
231
1
There are two parts here - first: energy is removed from the warm water to melt the ice. Adding this energy to the ice does not change the temperature of the ice - only the structure!

In the second part energy is removed from the warm water and added to the cold water (formerly the ice cube) until the two liquids are the same temperature.
 
  • #3
59
0
What I tryed is:

Water Q=mt = 0,3*4190 = 1257J

Ice to liquid: Q=mt = 0,15*334000 = 50100J

1) The melted Ice is now water of 0C = (0,15kg)*(4190)j/kg*K)(T-0)
2) The water of 50c = (0,3kg)*(4190)j/kg*K)(T-50)

solving that (0,15)*(4190)= 628,6T
(0,3)(4190)(T-50) = 1257T + 62850
628,5T+ 1257T=1885,5T

Just a shoot. Here to get a decent answer I took away the energy from the ice melted 62850-50100 = 12750/1885,5T = 6,76C

Comments ? :blushing:
 
  • #4
59
0
did I do something right here?
 
  • #5
59
0
Help :)
 
  • #6
59
0
This should be a easy task for you :p
 
  • #7
59
0
What I tryed is:

Water Q=mt = 0,3*4190 = 1257J

Ice to liquid: Q=mt = 0,15*334000 = 50100J

1) The melted Ice is now water of 0C = (0,15kg)*(4190)j/kg*K)(T-0)
2) The water of 50c = (0,3kg)*(4190)j/kg*K)(T-50)

solving that (0,15)*(4190)= 628,6T
(0,3)(4190)(T-50) = 1257T + 62850
628,5T+ 1257T=1885,5T

Just a shoot. Here to get a decent answer I took away the energy from the ice melted 62850-50100 = 12750/1885,5T = 6,76C

Comments ? :blushing:
any?
 
  • #8
59
0
why nobody ?
 
  • #9
alphysicist
Homework Helper
2,238
1
Hi Newton86,

Your work looks okay; but it's a bit hard to follow (especially how you subtracted the heat from the phase change). From your comments, I think it might help to see how it's organized. The formula would be:

(total heat gained) = (total heat lost)

which in this case is, since you know the ice will totally melt:

(heat to melt .15 kg ice) + (heat to warm .15 kg water) = (heat to cool .3 kg water)

or using your numbers

( .15 * 334000) + (.15 * 4190 * (T-0) ) = (.3 * 4190 * (50-T) )

But it looks like this is equivalent to what you did.
 
  • #10
59
0
Yeah actually :) Only I did not put them up that way
 

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