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Chemistry 15 Online
OpenStudy (anonymous):

Can someone check my specific heat answer?

OpenStudy (anonymous):

32.2-100.3=-68.1 C= -822.6j= (25.605)(-68.1) =2187.8j/(g x °C)

OpenStudy (anonymous):

@aaronq can you help me again?

OpenStudy (aaronq):

hm, you first need to find the heat exchanged. This is measured through the change in temperature of the water.

OpenStudy (anonymous):

Okay so would it be 25.3-32.2?

OpenStudy (anonymous):

6.9?

OpenStudy (aaronq):

it's actually 32.2-25.3, not the reverse. Because \(\Delta T=T_f-T_i\) (this is important because the sign confers where the energy is going) but yeah 6.9 celsius is right.

OpenStudy (anonymous):

221.6?

OpenStudy (aaronq):

i got 4.18 J/g/C*6.9 celsius *24 g*=692.208 J

OpenStudy (anonymous):

Okay you're probably right. Do you know how find out what kind of metal this is?

OpenStudy (anonymous):

I have to answer this question Using the experimental specific heat capacity value that you determined in Part II of the lab, what is the most probable identity of the metal that you examined?

OpenStudy (aaronq):

you can by identifying the specific heat capacity

OpenStudy (anonymous):

How do I do that?

OpenStudy (aaronq):

this is the exact the same problem you had previously. use \(q=m*C_p*\Delta T\)

OpenStudy (anonymous):

So the mass would be 25.605 right?

OpenStudy (anonymous):

692.208 would be the cp?

OpenStudy (anonymous):

and ΔT would be the change in temp but what two numbers would i use?

OpenStudy (anonymous):

@aaronq

OpenStudy (anonymous):

oh it would be 6.9!

OpenStudy (aaronq):

is the metal at an initial temp of 100.3 celsius?

OpenStudy (anonymous):

no it gets heated up to that

OpenStudy (aaronq):

by the water?

OpenStudy (anonymous):

yes

OpenStudy (aaronq):

i don't think thats right. i think the metal was introduced into the calorimeter at 100.3 celsius, then the temp decreased to 32.2 celsius.

OpenStudy (anonymous):

Okay you're probably right

OpenStudy (aaronq):

so you end up with \(692.208 ~J=(25.605~g)*C_p*(32.2-100.3)^{\circ}C\)

OpenStudy (anonymous):

what is the cp?

OpenStudy (aaronq):

specific heat capacity...

OpenStudy (anonymous):

and what is the number? i dont know what that is

OpenStudy (aaronq):

C is for capacity, the subscript P means that it's a value held at constant pressure

OpenStudy (aaronq):

use some algebra to find the value

OpenStudy (anonymous):

I'm very confused

OpenStudy (aaronq):

hm which part is not clear?

OpenStudy (anonymous):

I just need to figure out what the metal is

OpenStudy (aaronq):

did you find the specific heat capacity?

OpenStudy (anonymous):

No, i dont know the equation

OpenStudy (aaronq):

i wrote it a few posts ago. 692.208 J=(25.605 g)*Cp∗(32.2−100.3)∘C

OpenStudy (anonymous):

I got -1743.7

OpenStudy (aaronq):

damn, it should've been -692.208 J even so, your value should only be wrong by the sign. you mustve made a mistake

OpenStudy (aaronq):

i got 0.3969 J/g*C

OpenStudy (anonymous):

Okay and how do i find the metal using that?

OpenStudy (aaronq):

you need a table with their values to compare it. Did they give you one?

OpenStudy (anonymous):

OpenStudy (aaronq):

i would go with Ni because it's closest

OpenStudy (anonymous):

Okay thank you!

OpenStudy (aaronq):

no problem !

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