Heat Transfer and Temperature Equilibrium in a Mixed Substance System

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The discussion focuses on calculating the final temperature of a system consisting of mercury, water, and a brass calorimeter. The key equation used is Q=MC(deltaT), where heat lost by mercury equals heat gained by water and the calorimeter. Participants express confusion about determining the final temperature and setting up the equations correctly. They emphasize the need to isolate variables and rearrange the equation, particularly moving all terms involving final temperature (Tf) to one side. The conversation highlights the challenges of applying algebra in thermal equilibrium problems.
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?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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