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- anonymous

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)

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- anonymous

- schrodinger

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- Zale101

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

- Zale101

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

Mass is always in grams

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- Zale101

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

- Zale101

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

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

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