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 7845 J. What is the specific heat of the gas?

Answer 1

@Callisto

Answer 2

q=(m)(c)(delta T)

Answer 3

I know that i need to find c, but i dont know what is q in this case... o_0

Answer 4

\[7845j-346j=c(80.0g)(225C-25c)\]

Answer 5

.4686 <-- and when i entered it, it's wrong.. hmm?

Answer 6

it said "Check your value for q. When work is done by the system, the sign of w is negative. Sinc E=q w, a negative work value means that q is greater than E."

Answer 7

so would 7845/346=q?

Answer 8

let see

Answer 9

hmm that would result in .00141 as final answer...and it's wrong wth =.=

Answer 10

I'm guessing you are doing this on a computer program right?

Answer 11

yes..it's actually online homework..so each answer is checked after submit and it'll tell you right or wrong :D

Answer 12

i'm guessing it's a sig figs type thing then. I don't like these online homework things.

Answer 13

me too :(

Answer 14

and the worst thing is..when you get the problem wrong, it took off 5%. And you can't go back to re-do them neither... -.-

Answer 15

let me re-phrase that, each time you enter the wrong answer it took off 5% lol

Answer 16

work done by the system = 346J
Internal energy increased in the system = 7845J

Since Energy = Heat (q) + work (w),
7845J = q + (-346J).
Thus q = 8191J

And therefore, because Specific Heat = q/(mΔT),
c=8191J/(80.0g x (225℃ - 25℃))

I think that will do it.