Final Temperature of Copper and Water in Insulated Vessel

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Discussion Overview

The discussion revolves around a homework problem involving the final temperature of a system consisting of heated copper tubing placed in water within an insulated vessel. Participants are exploring the application of heat transfer principles and equations to solve for the final temperature, considering the contributions of both the copper and the water, as well as the heat capacity of the vessel.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant presents the initial setup of the problem, including the masses, specific heat capacities, and initial temperatures of the copper and water.
  • Another participant suggests assuming the initial temperature of the vessel is the same as that of the water, which is then incorporated into the equation.
  • A subsequent post questions the correctness of the signs in the heat transfer equation and discusses the need for consistency in the convention used for heat transfer calculations.
  • One participant clarifies that all heat changes should be on one side of the equation and mentions an alternative approach that groups heat lost and gained separately, emphasizing the importance of sticking to one convention.
  • There is a suggestion to solve the equation presented to check if the resulting temperature makes sense, indicating a focus on verifying the mathematical approach.

Areas of Agreement / Disagreement

Participants express differing views on the correct application of signs in the heat transfer equation and the conventions used for solving the problem. There is no consensus on the best approach, as some advocate for one method while others suggest alternatives.

Contextual Notes

Participants have not fully resolved the mathematical steps involved in the heat transfer calculations, and there are uncertainties regarding the application of conventions for heat transfer signs.

John Ker
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Homework Statement


A 529 g piece of copper tubing is heated to 89.5°C and placed in an insulated vessel containing 159 g of water at 22.8°C. Assuming no loss of water and a heat capacity for the vessel of 10.0 J/°C, what is the final temperature of the system [c of copper = 0.387 J/g · K]?

Homework Equations


q = q1 + q2 + q3 + ... = 0

Where q = (Tf - Ti) m c

The Attempt at a Solution



This is the part where I am stuck, knowing how to organize the equation.

Ive gotten this so far,

529 * .387 (Tf - 89.5) = 159 * 4.18 * (Tf - 22.8) + 10(Tf - Ti)

Im not sure how the vessel comes into this.

Thanks!
 
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Assume initial temperature of the vessel to be that of water it contained.
 
Borek said:
Assume initial temperature of the vessel to be that of water it contained.

SO the fomula would become:
529 * .387 (Tf - 89.5) = 159 * 4.18 * (Tf - 22.8) + 10(Tf - 22.8)

Where I then distribute everything out and solve for Tf?
Are the signs correct on both sides, I recall seeing that one side needs to be negative, but I am not sure that is relavent here.
 
No, signs are not OK. As your ΔT is defined as final-initial all heats (lost and gain) should be on one side of the equation, just like you wrote in your opening post.

It is also possible to solve such problems using a different convention, one in which we combine things that get colder (lose heat) on one side of the equation and things the gain heat (get hotter) on another side, then all changes in the temperature are assumed to be positive. Mathematically it is equivalent, can be sometimes easier conceptually. The only thing that really matters is that you stick to one convention and not mix them both.

In this particular case the error would be easy to spot - solve the equation you have listed for Tf, does the result make sense?
 
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