Finding entropy involving metal in water

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Homework Help Overview

The problem involves calculating the change in entropy for a thermally isolated system consisting of a piece of metal and water, where the metal is initially at a higher temperature than the water. The final temperature of the system is given, but the specific heat capacity and mass of the metal are unknown.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need to calculate heat transfer (Q) for both the metal and the water, questioning how to proceed without knowing the mass or type of metal.
  • Some participants suggest using the density of water to find its mass, while others express uncertainty about the feasibility of solving the problem without additional information.
  • There is mention of a potential formula for calculating entropy change, but participants are unsure about the implications of the process not being isothermal.

Discussion Status

The discussion is ongoing, with participants exploring different aspects of the problem. Some have suggested potential formulas and relationships between heat transfer and entropy, while others are still questioning the adequacy of the given information to arrive at a solution.

Contextual Notes

Participants note that the problem is part of an assignment, indicating a need for a solution despite the uncertainties regarding the metal's properties and the nature of the heat transfer process.

aal0315
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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?
 
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Please, can anybody help me with this question?
 
aal0315 said:
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.
 
Last edited:
I don't think it's possible without the metal :|
 
AtticusFinch said:
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?
 
aal0315 said:
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.
 
AtticusFinch said:
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.
 
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?
 
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.
 
  • #10
thank you for the help!
 
  • #11
AtticusFinch said:
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?
 

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