# Not sure if I calculated this correctly (Find the final temperature)

Did I solve this correctly? I’m not sure if my method is correct.

1. The problem statement, all variables and given/known data
A metal cylinder with the specific heat $c_{cylinder}$ and mass $m_{cylinder}$ is warmed up to temperature $T_{cylinder}$ and then cooled down by putting it into water which has the temperature $T_{water}$ and mass $m_{water}$ and specific heat $c_{water}$.
Find an expression for the final temperature $T$ (the equilibrium temperature).

Afterwards consider the following values:
$T_{cylinder}=600 °C$
$T_{water}=20 °C$
$m_{cylinder}=2 kg$
$m_{water}=2 kg$
$c_{water}=4190\frac{J}{kg\cdot K}$
$L_{water}=2256\frac{kJ}{kg}$ (heat of vaporization)

Calculate the final temperature if the cylinder was made of copper with specific heat $c_{Cu}=390\frac{J}{kg\cdot K}$ and comment on the result.

Then calculate the final temperature if the cylinder was made of aluminum with specific heat $c_{Al}=910\frac{J}{kg\cdot K}$ and comment on the result.

2. Relevant equations
Heat required for temperature change ΔT of mass m:
$Q=mcΔT$

Energy conservation relation:
$∑Q=0$

Heat transfer in a phase change:
$Q=mL$

3. The attempt at a solution
Using the first equation listed above for both the cylinder and water:
$Q_{cylinder}=m_{cylinder}c_{cylinder}(T-T_{cylinder})$
$Q_{water}=m_{water}c_{water}(T-T_{water})$

The energy conservation relation yields:
$Q_{cylinder}+Q_{water}=0$
$m_{cylinder}c_{cylinder}(T-T_{cylinder})+m_{water}c_{water}(T-T_{water})=0$

Isolating the final temperature T gives the expression:
$T=\frac{m_{water}c_{water}T_{water}+m_{cylinder}c_{cylinder}T_{cylinder }}{m_{water}c_{water}+m_{cylinder}c_{cylinder}}$

If the cylinder was made of copper:
I assume that the water does not boil so the final temperature will be less than 100 °C (boiling point of water).

$T=\frac{2\cdot 4190\cdot 20+2\cdot 390\cdot 600}{2\cdot 4190+2\cdot 390}=46 °C$

If the cylinder was made of aluminum:
I assume that the water does not boil so the final temperature will be less than 100 °C (boiling point of water).

$T=\frac{2\cdot 4190\cdot 20+2\cdot 910\cdot 600}{2\cdot 4190+2\cdot 910}=82 °C$

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