Latent heat and final temperature

[kimkibun]

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_H2O1=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_H2O1)=(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 iceQ_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..

 

[LawrenceC]

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.

 

[kimkibun]

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?

 

[LawrenceC]

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

Your answer is off by about 4.5 C. I have not gone through your calculations; I only did my own.

 

[kimkibun]

thank you very much sir! i finally got the correct answer. i made a wrong computation on the right hand side of the last equation. God bless!