1. The problem statement, all variables and given/known data 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 2. Relevant 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 3. 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..