# Latent heat and final temperature

## 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:

mAl=205g
mH2O=300g
TAl1=TH2O1=20°C
mFe=250g
mice=20g
TFe=150°C
Tice=-10°C

## Homework Equations

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

QAl+QH2O+Qice=QFe

Q=mcΔT

## The Attempt at a Solution

What is asked is the final temperature of mixture, Tmix. so,

QAl=(mAl)(cAl)(Tmix-TAl1)=(205g)(.21cal/g°C)(Tmix-20°C)=43.05Tmix-861

QH2O=(mH2O)(cH2O)(Tmix-TH2O1)=(300g)(1cal/g°C)(Tmix-20°C)=300Tmix-6000

Qice=(mice)(cice)(Tmix-Tice)+(mice)(Lf)=(20g)(.5cal/g°C)(Tmix-(-10°C))+(20g)(80cal/g)=10Tmix+100cal+1600cal

Lf is the latent heat of ice

QFe=(mFe)(cFe)(TFe-Tmix)=(250g)(.11cal/g°C)(150°C-Tmix)=4125cal-27.5Tmix

Therefore,

(43.05Tmix-861cal)+(300Tmix-6000cal)+(10Tmix+100cal+1600cal)=4125cal-27.5Tmix

380.55Tmix=7564cal

Tmix=19.88°C

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

Related Introductory Physics Homework Help News on Phys.org
Do this simple check to determine yourself whether your answer is correct. Take your final answer of 19.9 C and look at energy balances. You have energy lost by the vessel and its water. It is lost because the final temperature is lower than its initial temperature. Now determine how much energy the ice gained going from -10 C to 19.9 C including latent heat. Do the same for the energy loss of the iron going from 150 C to 19.9 C. Do you have a balance? If not, you did something wrong.

Do this simple check to determine yourself whether your answer is correct. Take your final answer of 19.9 C and look at energy balances. You have energy lost by the vessel and its water. It is lost because the final temperature is lower than its initial temperature. Now determine how much energy the ice gained going from -10 C to 19.9 C including latent heat. Do the same for the energy loss of the iron going from 150 C to 19.9 C. Do you have a balance? If not, you did something wrong.

i already check it. and yes it's wrong.

Since the heat gained by aluminum container, water, and ice is equal to heat lost by iron
is my assumption correct?

Yes, the heat gained by the aluminum container, water, and ice equals the heat lost by iron.