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

1. ### format1998

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

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 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

2. Relevant equations

Specific Heat Q = mcΔT
Latent Heat Q = mLf

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

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

2. ### technician

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

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

11,786
Re: Specific & Latent Heat - Heat required to turn water in Aluminum Tray --> Ice

I think you might omit 103 from the latent heat.

ehild

4. ### format1998

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

5. ### technician

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

There is certainly nothing wrong with your method (as far as I can see) it must be a computational error somewhere.

6. ### format1998

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

"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 $\frac{J}{kg*C°}$)(-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. ### format1998

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

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

11,786
Re: Specific & Latent Heat - Heat required to turn water in Aluminum Tray --> Ice

The latent heat has to be taken with negative sign. It is also removed heat. Change + to minus.

ehild

9. ### technician

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

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. ### format1998

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

Thank you both for your help!!! Much appreciated!