Calculating Final Temperature of Water and Iron Mixture

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Discussion Overview

The discussion revolves around calculating the final temperature of a mixture of water and iron after they reach thermal equilibrium. Participants explore the application of the principle of conservation of energy in the context of a homework problem involving specific heat capacities and temperature changes.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • The initial setup of the heat transfer equation is presented, where the heat lost by iron equals the heat gained by water.
  • One participant identifies a sign error in the original equation, suggesting that the temperature changes for iron and water should have different signs.
  • Another participant expresses confusion regarding the implications of the calculated temperature being lower than both initial temperatures.
  • A later reply emphasizes the importance of correctly applying the concept of heat flow, indicating that the heat gained and lost must be treated with opposite signs.
  • One participant attempts to correct their calculations by applying the concept of heat in and heat out, leading to a new temperature calculation.
  • Another participant clarifies that the change in temperature should reflect the direction of heat flow, reinforcing the need for consistent sign usage in the equations.
  • Participants express gratitude for the explanations and clarifications provided throughout the discussion.

Areas of Agreement / Disagreement

There is no consensus on the correct approach to the problem, as participants continue to discuss and refine their understanding of the equations and concepts involved. Disagreements remain regarding the proper treatment of temperature changes and signs in the calculations.

Contextual Notes

Participants have not fully resolved the mathematical steps involved in the calculations, and there are ongoing discussions about the assumptions related to heat transfer and temperature changes.

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Homework Statement



A piece of iron weighed at 120 grams at 95.0 degrees celsius is added to 750ml water of 25.0 degrees celsius. What is the final temperature of the water when it is equilibrated? Assume the dentisy of water is 1 g/ml. (Specific heat of water is 4.184 and specific heat of iron is .449).

Homework Equations



q=mc(delta t). In this case the q of iron is equal to the q of water.

The Attempt at a Solution



Here is my setup:

120*.449(x-95)=750*4.184(x-25)

Solving:
53.88x-5118.6=3138x-78450
53.88x=3138x-73331.4
-3084.12x=-73331.4
x=23.85

I know this is wrong, but I'm not sure if it's because I set this up wrong or did my calculations incorrectly.
 
Last edited:
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It's a sign error, the change in temperature from 95>x and 25->x must be different signs
 
I don't understand what you mean. I know that 23.85 implies that the temperature decreased for both the iron and water, and that is wrong.
 
mgb points out that there is a sign error in your original equation.

--
 
I think I forgot the qin=-qout concept. Let me try again.

120*.449(x-95)=-750*4.184(x-25)
53.88x-5118.6=-3138x+78450
3191.88x=83568.6

x=26.18 degrees celsius

Is this what you were referring to?
 
Almost,
if one thing is cooling from 95deg to x then the change in T is (95-x)
then if the other is heating form 25 to X the change must be (x-25) which gives the different signs.

Putting a minus on one side is the same thing (or at least has the same effect)
 
I thought that in order to do the equation correctly, one must always subtract the initial temperature from the final temperature.
 
The change in temperature defines the heat flow, so heat in and heat out would have different signs.
 
I see. Thank you for checking out my question and taking the time to explain it. :)
 

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