Equilibrium Temperature

  • Thread starter mikep
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  • #1
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A 199 g lead ball at a temperature of 80.9°C is placed in a light calorimeter containing 178 g of water at 24.5°C. Find the equilibrium temperature of the system.

i used the equation [tex]Q = mc \Delta T[/tex]
-(199g)(4.186J/g°C)(T - 80.9°C) + (178g)(0.13J/g°C)(T - 24.5°C) = 0
T = T final = 82.5°C
can someone please tell mw what i did wrong?
can someone also confirm if the specific heat for lead is 0.13J/g°C?
 

Answers and Replies

  • #2
Tom Mattson
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mikep said:
can someone please tell mw what i did wrong?

You put the specific heat of water in the term for the lead ball, and vice versa.

can someone also confirm if the specific heat for lead is 0.13J/g°C?

Not off the top of my head, but if you type "specific heat lead" into Google, I'm sure you can find it.
 
  • #3
43
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oh yeah and i also didn't need the negative sign. i got it thanks
 

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