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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?
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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whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.1\[\Delta Q = m c \Delta T\] 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
 one year ago
Best ResponseYou've already chosen the best response.0it says answer in 3 sig figs and i put 34.0... still wrong hmmm

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.1that's not what I get... show me your work?

marmar10
 one year ago
Best ResponseYou've already chosen the best response.0\[25.6= 18.4*.215*Tf22\] \[47.6= 3.956Tf\] Tf=12.032 12.032+22=34.0

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.1no...get that 22 out of there.

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.1change in heat = mass * c * change in temperature. starting temperature does not matter

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.1still 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
 one year ago
Best ResponseYou've already chosen the best response.1sorry, that is what I'm getting :)

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.1was looking at the wrong number on my scratch paper
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