Equilibrium Temperature with three substances

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Discussion Overview

The discussion revolves around calculating the equilibrium temperature of a system consisting of lead, aluminum, and water. Participants explore the application of specific heat equations and the principles of thermal equilibrium in a calorimetry context.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant presents a calculation for the equilibrium temperature using the formula Tf=((250x900x315)+(200x900x15)+(900x4.1x15))/((250x130)+(200x900)+(900x4.1), arriving at Tf=40.1, but expresses uncertainty about the correctness of their approach.
  • Another participant suggests a systematic approach to the problem, emphasizing the use of the equation \sum_i m_ic_i\Delta T_i = 0, where \Delta T_i is defined as T_{final} - T_{initial(i)}.
  • A question is raised regarding the units of specific heat for the substances involved, specifically noting the values for lead, aluminum, and water.
  • A repeated post reiterates the initial problem statement and calculations but points out inconsistencies with the second relevant equation, prompting a request for reevaluation.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the correct approach or the validity of the calculations presented. There are indications of confusion and requests for clarification, suggesting multiple competing views on the methodology.

Contextual Notes

The discussion includes potential limitations in the calculations, such as missing assumptions or errors in applying the relevant equations. The specific heat values and their implications for the calculations are also under scrutiny.

Who May Find This Useful

This discussion may be useful for students or individuals interested in thermodynamics, calorimetry, and the application of specific heat in solving thermal equilibrium problems.

ContagiousIntellect

Homework Statement


If a block of 250 grams of lead (specific heat (c) =130) at 315 degrees celsius is placed in a 200 gram aluminum (c=900) calorimeter cup containing 900 grams of water (c=4.1), and the calorimeter and the water are both initially at 15 degrees celsius, what is the equilibrium temperature ?

Homework Equations


Q=mcΔt
Tf=(mct+mct+mct)/(mc+mc+mc)

The Attempt at a Solution


Tf=((250x900x315)+(200x900x15)+(900x4.1x15))/((250x130)+(200x900)+(900x4.1)
Tf=(70875000+2700000+55350)/(32500+180000+3690)
Tf=40.1
I'm not sure if I'm on the right track.
 
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Try to approach it in a systematic way - for a single substance Q=mcΔt, for the whole system

\sum_i m_ic_i\Delta T_i = 0

where \Delta T_i = T_{final} -T_{initial(i)} (Tfinal is common for all substances present).
 
What are the units of specific heat, if lead is 130, aluminium is 900 and water is 4.1?
 
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ContagiousIntellect said:

Homework Statement


If a block of 250 grams of lead (specific heat (c) =130) at 315 degrees celsius is placed in a 200 gram aluminum (c=900) calorimeter cup containing 900 grams of water (c=4.1), and the calorimeter and the water are both initially at 15 degrees celsius, what is the equilibrium temperature ?

Homework Equations


Q=mcΔt
Tf=(mct+mct+mct)/(mc+mc+mc)

The Attempt at a Solution


Tf=((250x900x315)+(200x900x15)+(900x4.1x15))/((250x130)+(200x900)+(900x4.1)
Tf=(70875000+2700000+55350)/(32500+180000+3690)
Tf=40.1
I'm not sure if I'm on the right track.
The math is not consistent with your second Relevant equation. Try again.
 
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