Moderators: Chem_Mod, Chem_Admin

Alicia Lin 2F
Posts: 83
Joined: Wed Sep 18, 2019 12:17 am


Postby Alicia Lin 2F » Wed Jan 29, 2020 11:20 pm

The internal energy of a system increased by 982 J when it was supplied with 492 J of energy as heat.
(b) How much work was done?

Wouldn't you just calculate the work by taking the total energy of the system and subtracting the energy from heat (since internal energy is equal to heat plus work)?
When I do this, I get 49J, but the answer is 90x10^2 J.

Samuel G Rivera - Discussion 4I
Posts: 61
Joined: Sat Aug 17, 2019 12:16 am
Been upvoted: 1 time

Re: 4b.3b

Postby Samuel G Rivera - Discussion 4I » Thu Jan 30, 2020 1:52 am

You are right that you subtract the energy from heat (492 J) from the total energy (982 J) to get the measure of work. The answer should be +490 J. You probably made a small mistake when you calculated for this problem.

But you're right: delta U = w + q

Norman Dis4C
Posts: 101
Joined: Sat Sep 28, 2019 12:16 am

Re: 4b.3b

Postby Norman Dis4C » Sun Feb 02, 2020 1:23 pm

but why is the answer not 490J but 9000J?

Posts: 103
Joined: Thu Jul 11, 2019 12:17 am
Been upvoted: 1 time

Re: 4b.3b

Postby JohnWalkiewicz2J » Sun Feb 02, 2020 1:31 pm

Norman Dis4C wrote:but why is the answer not 490J but 9000J?

The answer is 4.90 x 10^2 J , which could also me written as "490 J" as the sig figs are the same. Im sure it was just a small miscalculation on his part.

Return to “Thermodynamic Systems (Open, Closed, Isolated)”

Who is online

Users browsing this forum: No registered users and 1 guest