Finding entropy involving metal in water

[aal0315]

Homework Statement

A piece of metal at 80C is placed into 1.2L of water at 72C. This thermally isolated system reaches a final temperature of 76C. Estimate the overall change of entropy for this system.

Homework Equations

deltaS = Q/T
Q=m*c*deltaT

The Attempt at a Solution

I think that i have to find Q for metal and water separately. correct? and if so i don't know how to find Q for the metal because i don't have its mass or what kind of metal it is to know its heat capacity.
does anyone have any suggestions?

 

[aal0315]

Please, can anybody help me with this question?

 

[AtticusFinch]

i don't know how to find Q for the metal because i don't have its mass or what kind of metal it is to know its heat capacity.
does anyone have any suggestions?

The amount of heat that leaves the metal is equal to the amount of heat that goes into the water.

 

[iRaid]

I don't think it's possible without the metal :|

 

[aal0315]

The amount of heat that leaves the metal is equal to the amount of heat that goes into the water.

But don't I still need to know the type of metal or mass of metal to find the Q in order to find the deltaS for the metal?

And I don't even know how to find out the water part because what is 1.2L in kg?

 

[AtticusFinch]

But don't I still need to know the type of metal or mass of metal to find the Q in order to find the deltaS for the metal?

And I don't even know how to find out the water part because what is 1.2L in kg?

Well you can find the mass of the water using the density of the water. But I just noticed this process isn't isothermal so I do think you are lacking information here.

 

[iRaid]

Well you can find the mass of the water using the density of the water. But I just noticed this process isn't isothermal so I do think you are lacking information here.

That's what I thought too.

 

[aal0315]

there has to be a way to answer this question .. it is part of an assignment that I have to hand in, so any other suggestions?

 

[AtticusFinch]

Thinking about it some more there is a solution.

You should be able to show using a formula from your book that

\Delta S = m_w c_w ln(\frac{349}{345}) + m_m c_m ln(\frac{349}{353})

Also, since heat in equals heat out

m_w c_w (349-345) = -m_m c_m (349-353)

That's all you need to solve this problem.

 

[aal0315]

thank you for the help!

 

[mlostrac]

Thinking about it some more there is a solution.

You should be able to show using a formula from your book that

\Delta S = m_w c_w ln(\frac{349}{345}) + m_m c_m ln(\frac{349}{353})

Also, since heat in equals heat out

m_w c_w (349-345) = -m_m c_m (349-353)

That's all you need to solve this problem.

I am having this same problem. What's the ln mean?