- #1

- 1,235

- 1

I know that

energy relased = s * m * ∆T

So would ∆T = 25 - 23.5?

Also would the mass = 28.2 g

Finally how would you calculate the specific heat capacity of only

**nickel**?

Thanks a lot

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter courtrigrad
- Start date

- #1

- 1,235

- 1

I know that

energy relased = s * m * ∆T

So would ∆T = 25 - 23.5?

Also would the mass = 28.2 g

Finally how would you calculate the specific heat capacity of only

Thanks a lot

- #2

Pyrrhus

Homework Helper

- 2,178

- 1

[tex] -Q_{nickel} = Q_{water} [/tex]

[tex] -m_{nickel}c_{nickel} (T_{final} - T_{nickel}) = m_{water}c_{water}(T_{final} - T_{water}) [/tex]

- #3

- 1,235

- 1

oh ok

so the specific heat of water = specific heat of nickel?

so the specific heat of water = specific heat of nickel?

- #4

Pyrrhus

Homework Helper

- 2,178

- 1

No, you need to work out the equation to get it.

- #5

- 1,235

- 1

wait what is c? aren't you supposed to use s * m * ∆T??

- #6

Pyrrhus

Homework Helper

- 2,178

- 1

C is specific heat. I apologize for my notation, that's how i remember it from the course.

- #7

- 1,235

- 1

- #8

Pyrrhus

Homework Helper

- 2,178

- 1

A more "savvy" way, i remember using was looking at the final temperature and at the initial temperature, as you can see the final temperature < initial temperature therefore the incrementum will be negative for the nickel.

- #9

- 1,235

- 1

hmmm got 0.93 when answer is 0.45. For ∆T of nickel and water, do you use the same final temperature 25.0 C?

Last edited:

- #10

Pyrrhus

Homework Helper

- 2,178

- 1

I got 0.445 j/g C, yes you do that's the whole point of thermo equilibrium.

- #11

- 1,235

- 1

yes i got that. thanks a lot for helping me

Share: