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## Homework Statement

An aluminum container whose mass is 205g contains 300g of water at 20°C. In this container is then placed 250g of iron at 150°C and 20g of ice at -10°C. Find the final temperature of the mixture.

Given:

m

_{Al}=205g

m

_{H2O}=300g

T

_{Al1}=T

_{H2O}

_{1}=20°C

m

_{Fe}=250g

m

_{ice}=20g

T

_{Fe}=150°C

T

_{ice}=-10°C

## Homework Equations

Since the heat gained by aluminum container, water, and ice is equal to heat lost by iron, then

Q

_{Al}+Q

_{H2O}+Q

_{ice}=Q

_{Fe}

Q=mcΔT

## The Attempt at a Solution

What is asked is the final temperature of mixture, T

_{mix}. so,

Q

_{Al}=(m

_{Al})(c

_{Al})(T

_{mix}-T

_{Al1})=(205g)(.21cal/g°C)(T

_{mix}-20°C)=43.05T

_{mix}-861

Q

_{H2O}=(m

_{H2O})(c

_{H2O})(T

_{mix}-T

_{H2O}

_{1})=(300g)(1cal/g°C)(T

_{mix}-20°C)=300T

_{mix}-6000

Q

_{ice}=(m

_{ice})(c

_{ice})(T

_{mix}-T

_{ice})+(m

_{ice})(L

_{f})=(20g)(.5cal/g°C)(T

_{mix}-(-10°C))+(20g)(80cal/g)=10T

_{mix}+100cal+1600cal

L

_{f}is the latent heat of ice

Q

_{Fe}=(m

_{Fe})(c

_{Fe})(T

_{Fe}-T

_{mix})=(250g)(.11cal/g°C)(150°C-T

_{mix})=4125cal-27.5T

_{mix}

Therefore,

(43.05T

_{mix}-861cal)+(300T

_{mix}-6000cal)+(10T

_{mix}+100cal+1600cal)=4125cal-27.5T

_{mix}

380.55T

_{mix}=7564cal

T

_{mix}=19.88°C

Is this correct? i think there's something wrong in my solution..