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

## 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 m

_{Al}= 340g

Water m

_{W}= 200g

T

_{i}= 18°C

T

_{f}= -15°C

## Homework Equations

Specific Heat Q = mcΔT

Latent Heat Q = mL

_{f}

## 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°C-18°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°C-0°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!