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Topic: Conservation Of Energy Final Temperature Of Thermal Equalibrium(Answer)  (Read 2871 times)

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Offline Jekel0000

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23) A 192-g piece of copper is heated to 100.0 °C in a boiling water bath and then dropped into a beaker containing 750 mL of water (density = 1.00 g/cm³) at 4.0 °C. What is the final temperature of the copper and water after they come to thermal equilibrium? (The specific heat of copper is 0.385 J/g•K). the specific heat of water is 4.184 J/g·K

the answer in the book is 6.22°C I believe it is because at the end of the equation they divided
96/42 then added 4 getting 6.22°C. I think it is wrong...

Offline Jekel0000

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In this problem qmetal is the amount of heat energy of a Metal
qmetal units are in Joules(kinetic energy) not calories(heat energy) and is equal to

qmetal =
(mass of the Metal)·(specific heat of the metal)·(Final Temperature - Initial Temperature)

specific heat is measured in Joule/grams·(change in temperature)
Temperature is Kelvin or Celsius
mass = weight/gravity....

Offline Borek

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6.2°C is the correct answer.

Solution you attached is probably the most convoluted thing I have seen in the last few years. You should plug expressions for qmetal and qwater into

[tex]q_{metal} = -q_{water}[/tex]

and solve directly for Tfinal.
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Offline Jekel0000

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Re: Conservation Of Energy Final Temperature Of Thermal Equalibrium(Answer)
« Reply #3 on: November 25, 2013, 06:50:38 AM »
Could you show me?

Offline Borek

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Re: Conservation Of Energy Final Temperature Of Thermal Equalibrium(Answer)
« Reply #4 on: November 25, 2013, 07:20:50 AM »
Yes. But I won't. Try to follow what you were told, that's the best way of learning.
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