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

by format1998
Tags: aluminum, heat, latent, required, specific, tray, turn, water
 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 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 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!
 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.
 HW Helper Thanks P: 10,673 I think you might omit 103 from the latent heat. ehild
 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.
 P: 1,506 There is certainly nothing wrong with your method (as far as I can see) it must be a computational error somewhere.
P: 26
 Quote by technician 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 $\frac{J}{kg*C°}$)(-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?
P: 26
 Quote by technician 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?
HW Helper
Thanks
P: 10,673
 Quote by format1998 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)
The latent heat has to be taken with negative sign. It is also removed heat. Change + to minus.

ehild
 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
 P: 26 Thank you both for your help!!! Much appreciated!

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