- #1

PhyIsOhSoHard

- 158

- 0

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

A metal cylinder with the specific heat [itex]c_{cylinder}[/itex] and mass [itex]m_{cylinder}[/itex] is warmed up to temperature [itex]T_{cylinder}[/itex] and then cooled down by putting it into water which has the temperature [itex]T_{water}[/itex] and mass [itex]m_{water}[/itex] and specific heat [itex]c_{water}[/itex].

Find an expression for the final temperature [itex]T[/itex] (the equilibrium temperature).

Afterwards consider the following values:

[itex]T_{cylinder}=600 °C[/itex]

[itex]T_{water}=20 °C[/itex]

[itex]m_{cylinder}=2 kg[/itex]

[itex]m_{water}=2 kg[/itex]

[itex]c_{water}=4190\frac{J}{kg\cdot K}[/itex]

[itex]L_{water}=2256\frac{kJ}{kg}[/itex] (heat of vaporization)

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

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

Heat required for temperature change ΔT of mass m:

[itex]Q=mcΔT[/itex]

Energy conservation relation:

[itex]∑Q=0[/itex]

Heat transfer in a phase change:

[itex]Q=mL[/itex]

Using the first equation listed above for both the cylinder and water:

[itex]Q_{cylinder}=m_{cylinder}c_{cylinder}(T-T_{cylinder})[/itex]

[itex]Q_{water}=m_{water}c_{water}(T-T_{water})[/itex]

The energy conservation relation yields:

[itex]Q_{cylinder}+Q_{water}=0[/itex]

[itex]m_{cylinder}c_{cylinder}(T-T_{cylinder})+m_{water}c_{water}(T-T_{water})=0[/itex]

Isolating the final temperature T gives the expression:

[itex]T=\frac{m_{water}c_{water}T_{water}+m_{cylinder}c_{cylinder}T_{cylinder}}{m_{water}c_{water}+m_{cylinder}c_{cylinder}}[/itex]

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).

[itex]T=\frac{2\cdot 4190\cdot 20+2\cdot 390\cdot 600}{2\cdot 4190+2\cdot 390}=46 °C[/itex]

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).

[itex]T=\frac{2\cdot 4190\cdot 20+2\cdot 910\cdot 600}{2\cdot 4190+2\cdot 910}=82 °C[/itex]

## Homework Statement

A metal cylinder with the specific heat [itex]c_{cylinder}[/itex] and mass [itex]m_{cylinder}[/itex] is warmed up to temperature [itex]T_{cylinder}[/itex] and then cooled down by putting it into water which has the temperature [itex]T_{water}[/itex] and mass [itex]m_{water}[/itex] and specific heat [itex]c_{water}[/itex].

Find an expression for the final temperature [itex]T[/itex] (the equilibrium temperature).

Afterwards consider the following values:

[itex]T_{cylinder}=600 °C[/itex]

[itex]T_{water}=20 °C[/itex]

[itex]m_{cylinder}=2 kg[/itex]

[itex]m_{water}=2 kg[/itex]

[itex]c_{water}=4190\frac{J}{kg\cdot K}[/itex]

[itex]L_{water}=2256\frac{kJ}{kg}[/itex] (heat of vaporization)

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

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

## Homework Equations

Heat required for temperature change ΔT of mass m:

[itex]Q=mcΔT[/itex]

Energy conservation relation:

[itex]∑Q=0[/itex]

Heat transfer in a phase change:

[itex]Q=mL[/itex]

## The Attempt at a Solution

Using the first equation listed above for both the cylinder and water:

[itex]Q_{cylinder}=m_{cylinder}c_{cylinder}(T-T_{cylinder})[/itex]

[itex]Q_{water}=m_{water}c_{water}(T-T_{water})[/itex]

The energy conservation relation yields:

[itex]Q_{cylinder}+Q_{water}=0[/itex]

[itex]m_{cylinder}c_{cylinder}(T-T_{cylinder})+m_{water}c_{water}(T-T_{water})=0[/itex]

Isolating the final temperature T gives the expression:

[itex]T=\frac{m_{water}c_{water}T_{water}+m_{cylinder}c_{cylinder}T_{cylinder}}{m_{water}c_{water}+m_{cylinder}c_{cylinder}}[/itex]

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).

[itex]T=\frac{2\cdot 4190\cdot 20+2\cdot 390\cdot 600}{2\cdot 4190+2\cdot 390}=46 °C[/itex]

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).

[itex]T=\frac{2\cdot 4190\cdot 20+2\cdot 910\cdot 600}{2\cdot 4190+2\cdot 910}=82 °C[/itex]

Last edited: