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

• PhyIsOhSoHard
In summary, we are given a problem where a metal cylinder with a specific heat c_{cylinder}, mass m_{cylinder}, and temperature T_{cylinder} is cooled down by being placed in water with a specific heat c_{water}, mass m_{water}, and temperature T_{water}. We are asked to find the final temperature T at equilibrium. By using the equations for heat required for temperature change, energy conservation, and heat transfer during a phase change, we can derive an expression for T. By plugging in the given values and assuming that the water does not boil, we can find that the final temperature is 69 °C if the cylinder is made of copper and 123 °C if the cylinder is made of
PhyIsOhSoHard
Did I solve this correctly? I'm not sure if my method is correct.

## Homework Statement

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.

## Homework Equations

Heat required for temperature change ΔT of mass m:
$Q=mcΔT$

Energy conservation relation:
$∑Q=0$

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

## 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:
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.

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

1 person
Chestermiller said:
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

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?

PhyIsOhSoHard said:
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?
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

1 person

## 1. How do I know if I calculated the final temperature correctly?

To verify your calculation, you can compare it with the expected result or use a formula or calculator to double check your work. You can also ask a colleague or a mentor to review your calculation.

## 2. What factors should I consider when calculating the final temperature?

The initial temperature, the amount of heat added or removed, the specific heat capacity of the substance, and any phase changes (such as melting or boiling) are important factors to consider when calculating the final temperature.

## 3. Can I use any formula to calculate the final temperature?

No, the formula used to calculate the final temperature depends on the specific scenario and the properties of the substances involved. It is important to use the correct formula and units to get an accurate result.

## 4. What are some common mistakes when calculating the final temperature?

Some common mistakes include using the wrong formula, using incorrect units, not accounting for phase changes, and not considering the specific heat capacity of the substances involved. It is important to double check your work and ensure all factors are accounted for in the calculation.

## 5. How can I improve my accuracy when calculating the final temperature?

To improve accuracy, make sure you have all the necessary information and use the correct formula and units. It is also helpful to round off numbers only at the end of the calculation, and to double check your work using a calculator or by hand. Additionally, practicing and familiarizing yourself with the concepts and formulas can also improve accuracy.

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