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Specific & Latent Heat  Heat required to turn water in Aluminum Tray > Ice 
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#1
Nov811, 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 18°C. How much heat (Q) must be removed to turn the water into ice at 15°C? Aluminum m_{Al}= 340g Water m_{W}= 200g T_{i}= 18°C T_{f}= 15°C 2. Relevant equations Specific Heat Q = mcΔT Latent Heat Q = mL_{f} 3. The attempt at a solution Q_{NET}= heat removed to bring Aluminum from 18°C to 15°C + heat removed to bring Water_{l} 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 Q_{NET} = (mcΔT)_{Al} + (mcΔT)_{W(l}) + (mL_{f})_{w} + (mcΔT)_{W(s}) Q_{NET} = [.34kg (900 [itex]\frac{J}{kg*°C}[/itex])(15°C18°C)] + [.2kg (4186 [itex]\frac{J}{kg*°C}[/itex])(0°C  18°C)] +[.2kg (333*10^{3} [itex]\frac{J}{kg}[/itex])] + [.2kg (2100 [itex]\frac{J}{kg*°C}[/itex]) (15°C0°C) Q_{NET} = 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! 


#2
Nov811, 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. 


#3
Nov811, 02:46 PM

HW Helper
Thanks
P: 10,551

I think you might omit 10^{3} from the latent heat.
ehild 


#4
Nov811, 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 10^{3} 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*10^{3} J/kg. 


#5
Nov811, 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.



#6
Nov811, 03:06 PM

P: 26

Q = mcΔT 33 grams for the mass of the Aluminum tray? Q_{Al}= (.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
Nov811, 03:13 PM

P: 26

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
Nov811, 03:21 PM

HW Helper
Thanks
P: 10,551

ehild 


#9
Nov811, 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 


#10
Nov811, 04:25 PM

P: 26

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



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