Specific & Latent Heat - Heat required to turn water in Aluminum Tray -> Ice

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Specific & Latent Heat - Heat required to turn water in Aluminum Tray --> Ice

Homework Statement



200 g Water (l) is contained in an Aluminum ice tray that has a mass of 340 g. Both is at 18°C. How much heat (Q) must be removed to turn the water into ice at -15°C?

Aluminum mAl= 340g
Water mW= 200g
Ti= 18°C

Tf= -15°C


Homework Equations



Specific Heat Q = mcΔT
Latent Heat Q = mLf

The Attempt at a Solution



QNET= heat removed to bring Aluminum from 18°C to -15°C
+ heat removed to bring Waterl from 18°C to 0°C
+ heat removed from water to change phase (liquid to solid)
+ heat removed to bring ice from 0°C to -15°C

QNET = (mcΔT)Al + (mcΔT)W(l) + (mLf)w + (mcΔT)W(s)

QNET = [.34kg (900 [itex]\frac{J}{kg*°C}[/itex])(-15°C-18°C)] + [.2kg (4186 [itex]\frac{J}{kg*°C}[/itex])(0°C - 18°C)] +[.2kg (333*103 [itex]\frac{J}{kg}[/itex])] + [.2kg (2100 [itex]\frac{J}{kg*°C}[/itex]) (-15°C-0°C)
QNET = 35132.4 J = 35.1324 kJ

My answer is not one of the choices. What am I doing wrong? Please help!

Thank you in advance. Any and all help is much appreciated!
 

Answers and Replies

  • #2
1,506
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when I did it I got 97.9kJ. I can only imagine that something went wrong with all the - signs.
For the Al I used temp change = 33 (18 to -15), for the water 18 and for the ice 15.
 
  • #3
ehild
Homework Helper
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I think you might omit 103 from the latent heat.

ehild
 
  • #4
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I think you might omit 103 from the latent heat.

Latent Heat as given on the table was 333 kJ/kg. I needed it to be J/kg, which came out to be 333*103 J/kg.
 
  • #5
1,506
18


There is certainly nothing wrong with your method (as far as I can see) it must be a computational error somewhere.
 
  • #6
26
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when I did it I got 97.9kJ. I can only imagine that something went wrong with all the - signs.
For the Al I used temp change = 33 (18 to -15), for the water 18 and for the ice 15.
"For the Al I used temp change = 33 (18 to -15)"

Q = mcΔT
33 grams for the mass of the Aluminum tray?
QAl= (.34 kg) (900 [itex]\frac{J}{kg*C°}[/itex])(-15°C - 18°C)
did I lay that out wrong or did I use the wrong numbers?

"for the water 18"
Water in liquid form is from 18°C to 0°C

"ice 15"
Ice from 0°C to -15°C

is that what you meant?
 
  • #7
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There is certainly nothing wrong with your method (as far as I can see) it must be a computational error somewhere.
Q(aluminum) = -10098 J
Q (water from 18°C to 0°C) = - 150696 J
Q (water to ice phase change) = 666000 J
Q (ice from 0°C to -15°C) = -6300 J

(-10098 J) + (-150696 J) + (666000 J) + (-6300 J) = 35,132.4 J = 35.13 kJ

What am I doing wrong? Are the signs wrong?
 
  • #8
ehild
Homework Helper
15,543
1,909


QNET = [.34kg (900 [itex]\frac{J}{kg*°C}[/itex])(-15°C-18°C)] + [.2kg (4186 [itex]\frac{J}{kg*°C}[/itex])(0°C - 18°C)] +[.2kg (333*103 [itex]\frac{J}{kg}[/itex])] + [.2kg (2100 [itex]\frac{J}{kg*°C}[/itex]) (-15°C-0°C)
The latent heat has to be taken with negative sign. It is also removed heat. Change + to minus.

ehild
 
  • #9
1,506
18


my values are
Q for aluminium =0.34 x 900 x 33 = 10098 J
Q water 18 to 0 = 0.2 x 4186 x 18 = 15070 J
Q water to ice = 0.2 x 333000 = 66600 J I think this is the difference!!!
Q ice to -15 = 0.2 x 2100 x 15 = 6300 J
These all add up to 10098 + 15070 +66600 +6300 = 98068 or 98kJ
We must be getting close !!!! I want to go to bed soon
 
  • #10
26
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Thank you both for your help!!! Much appreciated!
 

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