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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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
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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.
34 C?
it says answer in 3 sig figs and i put 34.0... still wrong hmmm

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

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