## A community for students. Sign up today

Here's the question you clicked on:

## 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?

• This Question is Closed
1. 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.

2. marmar10

34 C?

3. marmar10

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

4. whpalmer4

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

5. marmar10

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

6. whpalmer4

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

7. whpalmer4

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

8. marmar10

ohhhh

9. marmar10

so i got 28.5 now

10. 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}}}$

11. whpalmer4

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

12. whpalmer4

was looking at the wrong number on my scratch paper

13. marmar10

awesome thank you!

#### Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy