A 40.6 g sample of iron metal is heated and put into 100 g of water at 19.7 degrees Celsius in a calorimeter. If the final Temperature of the iron sample and the water is 24.3 degrees Celsius, what was the temperature of the iron sample when it was placed in the water?(I'll give a medal)

Question

zale101 · Answer

$\LARGE -q_{~lost ~by ~iron} = q_{~gained~by~ water} $

$~~~~~~~~~\large \color{purple}{-m*C*\Delta T =m*C*\Delta T} $

zale101 · Answer

m represents the mass, C represents the specific heat capacity, delta T represents the temperature difference/change 
Keep in mind that $\large \Delta T=T_{final}−T_{initial}$
 You are looking for T initial of iron and the rest is giving to you. Plug everything and solve T initial of Fe algebraically. 

                    −m∗C∗ΔT=m∗C∗ΔT
  $\large −m_{Fe}∗C_{Fe}∗(T_f-\color{red}{T_i})=m_{H_2O}∗C_{H_2O}∗(T_f-T_i)$

zale101 · Answer

Mass is always in grams

zale101 · Answer

and again, it 's already been given to you for both iron and water :)

zale101 · Answer

Specific heat capacity should be given to you
specific heat C. for  iron = 0.44 J/g°C
specific heat C. for  water = 4.18 J/g°C

zale101 · Answer

Temperatures are already given to you, expect for T initial for iron and that's what you are solving for.