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

1. Mar 18, 2014

### PhyIsOhSoHard

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).

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

Last edited: Mar 18, 2014
2. Mar 18, 2014

### PhyIsOhSoHard

I just realized that I miscalculated on both equations. Let me try again:

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).

$T=\frac{2\cdot 4190\cdot 20+2\cdot 390\cdot 600}{2\cdot 4190+2\cdot 390}=69 °C$

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).

$T=\frac{2\cdot 4190\cdot 20+2\cdot 910\cdot 600}{2\cdot 4190+2\cdot 910}=123.5 °C$

Since the final temperature is above 100 °C, that means the water boils and we have a phase change.

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

Isolating T:
$T=\frac{m_{water}c_{water}T_{water}+m_{cylinder}c_{cylinder}T_{cylinder}-L_{water}m_{water}}{m_{water}c_{water}+m_{cylinder}c_{cylinder}}$

Inserting values:
$T=\frac{2\cdot 4190\cdot 20+2\cdot 910\cdot 600-2256\cdot 2}{2\cdot 4190+2\cdot 910}=123 °C$

So the final temperature is 123 °C.

3. Mar 18, 2014

### Staff: Mentor

Try again. That's 2256.2 kJ/kg, not J/kg. Also, maybe not all the water vaporizes, in which case the final temperature would be 100C, and you would have to find out how much liquid water remains.

Chet

4. Mar 18, 2014

### PhyIsOhSoHard

Oh yes, you're right. Then I get:
$T=\frac{2\cdot 4190\cdot 20+2\cdot 910\cdot 600-2256000\cdot 2}{2\cdot 4190+2\cdot 910}=-319 °C$

That's clearly not correct though so I'm assuming this isn't the correct method that I'm using.

How do you know the final temperature is 100 °C? Is it because it can't get any higher than 100?

If we want to calculate how much water liquid remains we need to figure out how much x amount is vaporized $xm_{water}L_{water}$ at 100 degrees:

$m_{Al}c_{Al}(100-T_{Al})+xm_{water}L_{water}+m_{water}c_{water}(100-T_{water})=0$

Solve for x:
$x=\frac{-m_{Al}c_{Al}(100-T_{Al})-m_{water}c_{water}(100-T_{water})}{m_{water}L_{water}}$

Insert values:
$x=\frac{-2\cdot 910\cdot (100-600)-2\cdot 4190\cdot (100-20)}{2\cdot 2256000}=0.053$

That means a total of $0.053\cdot 2kg=0.106kg$ water is vaporized. So the water that remains at 100 degrees is:
$2kg-0.106kg=1.894kg$

But what now?

5. Mar 18, 2014

### Staff: Mentor

I haven't checked your arithmetic, but this is the right idea. The temperature would stay at 100C until, if necessary, all the water has been vaporized, which is not the case in this problem.

As far as "but what now?", nothing. You're done (if you did the math correctly).

Chet