Marmar10
  • Marmar10
Aluminum has a specific heat of 0.215 cal/(g⋅∘C). When 25.6 cal of heat is added to 18.4 g of aluminum at 22.0 ∘C, what is the final temperature of the aluminum?
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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whpalmer4
  • whpalmer4
\[\Delta Q = m c \Delta T\] solve that for \(\Delta T\) and use it to find the change in temperature, which you then add to the starting temperature to find the final temperature.
Marmar10
  • Marmar10
34 C?
Marmar10
  • Marmar10
it says answer in 3 sig figs and i put 34.0... still wrong hmmm

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whpalmer4
  • whpalmer4
that's not what I get... show me your work?
Marmar10
  • Marmar10
\[25.6= 18.4*.215*Tf-22\] \[47.6= 3.956Tf\] Tf=12.032 12.032+22=34.0
whpalmer4
  • whpalmer4
no...get that -22 out of there.
whpalmer4
  • whpalmer4
change in heat = mass * c * change in temperature. starting temperature does not matter
Marmar10
  • Marmar10
ohhhh
Marmar10
  • Marmar10
so i got 28.5 now
whpalmer4
  • whpalmer4
still not what I'm getting, but me check here... \[\Delta Q = mc\Delta T\]\[\Delta T = \frac{\Delta Q}{m c} = \frac{25.6 \text{ cal}}{18.4 \text{ g} * 0.215 \text{ cal/g }{^\circ \text{C}}}\]
whpalmer4
  • whpalmer4
sorry, that is what I'm getting :-)
whpalmer4
  • whpalmer4
was looking at the wrong number on my scratch paper
Marmar10
  • Marmar10
awesome thank you!

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