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?

Question

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?

whpalmer4 · Answer

$$\Delta Q = m c \Delta T$$
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 · Answer

34 C?

marmar10 · Answer

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

whpalmer4 · Answer

that's not what I get...
show me your work?

marmar10 · Answer

$$25.6= 18.4*.215*Tf-22$$
$$47.6= 3.956Tf$$
Tf=12.032
12.032+22=34.0

whpalmer4 · Answer

no...get that -22 out of there.

whpalmer4 · Answer

change in heat = mass * c * change in temperature.  starting temperature does not matter

marmar10 · Answer

ohhhh

marmar10 · Answer

so i got 28.5 now

whpalmer4 · Answer

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 · Answer

sorry, that is what I'm getting :-)

whpalmer4 · Answer

was looking at the wrong number on my scratch paper

marmar10 · Answer

awesome thank you!