- #1
format1998
- 26
- 0
Specific & Latent Heat - Heat required to turn water in Aluminum Tray --> Ice
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
Specific Heat Q = mcΔT
Latent Heat Q = mLf
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!
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!