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


by format1998
Tags: aluminum, heat, latent, required, specific, tray, turn, water
format1998
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#1
Nov8-11, 02:10 PM
P: 26
1. The problem statement, all variables and given/known data

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

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

Tf= -15C


2. Relevant equations

Specific Heat Q = mcΔT
Latent Heat Q = mLf

3. The attempt at a solution

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

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

QNET = [.34kg (900 [itex]\frac{J}{kg*C}[/itex])(-15C-18C)] + [.2kg (4186 [itex]\frac{J}{kg*C}[/itex])(0C - 18C)] +[.2kg (333*103 [itex]\frac{J}{kg}[/itex])] + [.2kg (2100 [itex]\frac{J}{kg*C}[/itex]) (-15C-0C)
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!
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technician
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#2
Nov8-11, 02:30 PM
P: 1,506
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.
ehild
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#3
Nov8-11, 02:46 PM
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P: 9,834
I think you might omit 103 from the latent heat.

ehild

format1998
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#4
Nov8-11, 02:59 PM
P: 26

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


[QUOTE=ehild;3605531]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.
technician
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#5
Nov8-11, 03:05 PM
P: 1,506
There is certainly nothing wrong with your method (as far as I can see) it must be a computational error somewhere.
format1998
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#6
Nov8-11, 03:06 PM
P: 26
Quote Quote by technician View Post
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])(-15C - 18C)
did I lay that out wrong or did I use the wrong numbers?

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

"ice 15"
Ice from 0C to -15C

is that what you meant?
format1998
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#7
Nov8-11, 03:13 PM
P: 26
Quote Quote by technician View Post
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 18C to 0C) = - 150696 J
Q (water to ice phase change) = 666000 J
Q (ice from 0C to -15C) = -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?
ehild
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#8
Nov8-11, 03:21 PM
HW Helper
Thanks
P: 9,834
Quote Quote by format1998 View Post
QNET = [.34kg (900 [itex]\frac{J}{kg*C}[/itex])(-15C-18C)] + [.2kg (4186 [itex]\frac{J}{kg*C}[/itex])(0C - 18C)] +[.2kg (333*103 [itex]\frac{J}{kg}[/itex])] + [.2kg (2100 [itex]\frac{J}{kg*C}[/itex]) (-15C-0C)
The latent heat has to be taken with negative sign. It is also removed heat. Change + to minus.

ehild
technician
technician is offline
#9
Nov8-11, 03:32 PM
P: 1,506
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
format1998
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#10
Nov8-11, 04:25 PM
P: 26
Thank you both for your help!!! Much appreciated!


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