Question

An 80.0-gram sample of a gas was heated from 25 °C to 225 °C. During this process, 346 J of work was done by the system and its internal energy increased by 8625 J. What is the specific heat of the gas? I keep getting .517J but the homework online is telling me I'm wrong. I really need to get this answer and understand where I went wrong! Thank you in advance!

Answer 1

By the first law of thermodynamics:
\[\Delta U = q + w\]

q is heat energy
w is work energy
U is the change in internal energy

Note:
Work done by the system is negative
Work done on the system is positive

The question gives you two of the variables so you can solve for q

Now we need the equation for heat capacity

\[C = \frac{q}{m*\Delta T}\]

where,

m = mass in grams
q = heat energy
C = cpecific heat capacity
\[\Delta T = T_{final\ Temperature} - T_{initial\ Temperature} = Change\ In\ Temperature\]

Solve for C and you are done

Answer 2

Remember temperature is in kelvin but it doesnt really matter since the 273 will cancel out anyways when you find the change

Answer 3

@Australopithecus So I solved the way you demonstrated using my values and I got a new calculation of .4629 J... Does this value seem correct?

Answer 4

Did you read everything I wrote?

Answer 5

I think you made a mistake

Answer 6

Show me your work? I can tell you where you are making a mistake

Answer 7

@Australopithecus So I tried again and I got the same answer as I did in the beginning which according to my online homework is wrong.
Using equation #1.)
200degrees=q+8971J
q+-8771J
Equation #2.)
C=-8771/((80)(9200degrees)) = -.548187 J What am I doing wrong?! ):

Answer 8

deltaU is the change in internal energy, so
\[\Delta U = q + w\]
\[8625=q+346\]

Answer 9

Does that step make sense?

Answer 10

Because it says that the change in internal energy is 8625 J and the work done is 346 J

Answer 11

@JoannaBlackwelder okay so following through with that, I get that q=8279 then if i place that into the other equation
8279/((80grams)(200deltaT)) then I get that c=.517 which is as well wrong in accordance to my online homework!

Answer 12

You put before that the unit is J, which is incorrect. What unit did you use?

Answer 13

It is work done by the system so work is negative

Answer 14

(J/g*C)=.517

Answer 15

Oh, and I just realized that it says that work was done by the system.

Answer 16

that is his mistake I believe

Answer 17

Yep, I think you are right.

Answer 18

8625 = q + (-346)
q = 8625 + 346

Answer 19

The mistake was in calculating q

Answer 20

so then q=8971, 8971/((80)(200))=C, C=.5606 (J/g*C)

Answer 21

or should that be a negative value because it is work done by the system?

Answer 22

it would be Joules/K*g

Answer 23

Remember when doing chemistry questions always convert to Kelvin unless otherwise told

Answer 24

but in this case it wont matter because 273 is canceled out but it is still converted

T = (225+273)K - (25 + 273)K

Answer 25

= 200K

Answer 26

The question wants it in (J/g*C) But im not sure if that makes a difference in this question

Answer 27

oh then it will be the same

Answer 28

so then the final answer would be (J/g*C)=-.56068 (negative not positive)?

Answer 29

you shouldn't be getting a negative

Answer 30

oh yeah you're right I'm just getting positive .56068

Answer 31

http://www.wolframalpha.com/input/?i=%288625%2B346%29%2F%28%28225+-+25%29*80%29

Answer 32

that is what I got

Answer 33

AND THE ANSWER IS CORRECT!!! Thank you so much for breaking it down for me!!! :D

Answer 34

No problem sorry for taking so long to get back to you

Answer 35

It's fine I appreciate your time ^^