Heat Transfer and Temperature Equilibrium in a Mixed Substance System

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Homework Help Overview

The problem involves heat transfer between a piece of mercury, water, and a brass calorimeter, with the goal of determining the final temperature of the mixture. The context is centered around thermodynamics and heat exchange principles.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the heat transfer equation Q=MC(deltaT) and the concept of heat lost equating to heat gained. There is confusion regarding the initial and final temperatures of the substances involved, particularly the mercury.

Discussion Status

Some participants are attempting to formulate expressions based on the heat transfer principles, while others express difficulty in manipulating the algebra involved. There is an ongoing exploration of how to set up the equations correctly, with no clear consensus reached yet.

Contextual Notes

Participants mention their educational backgrounds in algebra and precalculus, indicating a potential gap in confidence with the algebraic manipulation required to solve the problem. The initial temperatures of the substances are noted, but there is uncertainty about how to proceed with the calculations.

balllla
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Homework Statement



A .4lb piece of Hg at 210 F is placed in .3lb of water that has a temperature of 70.0f. The water is in .1lb brass calorimeter.

What is the final temperature of the mixture?

Homework Equations



Q=MC(deltaT)


The Attempt at a Solution



The part that I'm confused from is.

Qlost=Qgained
What is the final and initial temperature of the silver?
I can't figure that out from the problem, would the final temperature of the silver be 70.0degreeS?
 
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balllla said:
Qlost=Qgained
What is the final and initial temperature of the silver?
I can't figure that out from the problem, would the final temperature of the silver be 70.0degreeS?

The heat lost by the Hg = heat gained by the water + heat gained by the calorimeter

Initially, the water and calorimeter should be at the same temperature.

So can you formulate an expression for the given terms?
 
I cannot formulate an expression for the given terms. I've completed algebra, adv. algebra and now I am precalc and I just can't figure out the algebra to complete this practice problem.
 
balllla said:
I cannot formulate an expression for the given terms. I've completed algebra, adv. algebra and now I am precalc and I just can't figure out the algebra to complete this practice problem.

Let's call the final temperature Tf


using Q=McΔT

What is the heat lost from the Hg?
 
Well I understand that -(Qloss)=Qgained and you need to get all variables to one side and other numbers on the other side, this is what i have done but I'm stuck

-(Qloss)=Qgained
-(Qloss silver)=Qgained of water=Qgainedofcalirmiter
-(mcdeltat)=mcdeltat+mcdeltat
-(181.8181g(.214cal/gc)(tf-98.8c)=(136.3636g(1cal)(tf-21.0c)+(45.454545g)(.0917cal/gc)(tf-21.0c)

okay this is what I have now I do not understand the algebra. ?
 
Can you expand the brackets and then move all the Tf to one side?
 

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