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Marmar10

  • one year ago

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?

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  1. whpalmer4
    • one year ago
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    \[\Delta Q = m c \Delta T\] 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.

  2. marmar10
    • one year ago
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    34 C?

  3. marmar10
    • one year ago
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    it says answer in 3 sig figs and i put 34.0... still wrong hmmm

  4. whpalmer4
    • one year ago
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    that's not what I get... show me your work?

  5. marmar10
    • one year ago
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    \[25.6= 18.4*.215*Tf-22\] \[47.6= 3.956Tf\] Tf=12.032 12.032+22=34.0

  6. whpalmer4
    • one year ago
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    no...get that -22 out of there.

  7. whpalmer4
    • one year ago
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    change in heat = mass * c * change in temperature. starting temperature does not matter

  8. marmar10
    • one year ago
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    ohhhh

  9. marmar10
    • one year ago
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    so i got 28.5 now

  10. whpalmer4
    • one year ago
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    still not what I'm getting, but me check here... \[\Delta Q = mc\Delta T\]\[\Delta T = \frac{\Delta Q}{m c} = \frac{25.6 \text{ cal}}{18.4 \text{ g} * 0.215 \text{ cal/g }{^\circ \text{C}}}\]

  11. whpalmer4
    • one year ago
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    sorry, that is what I'm getting :-)

  12. whpalmer4
    • one year ago
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    was looking at the wrong number on my scratch paper

  13. marmar10
    • one year ago
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    awesome thank you!

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