# Specific Heat Problem Help!

## 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?

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 cant figure that out from the problem, would the final temperature of the silver be 70.0degreeS?

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rock.freak667
Homework Helper
Qlost=Qgained
What is the final and initial temperature of the silver?
I cant 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 im precalc and I just cant figure out the algebra to complete this practice problem.

rock.freak667
Homework Helper
I cannot formulate an expression for the given terms. I've completed algebra, adv. algebra and now im precalc and I just cant 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. ?

rock.freak667
Homework Helper
Can you expand the brackets and then move all the Tf to one side?