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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Zoloft on February 02, 2009, 01:08:28 AM

Title: Calorimeter again!
Post by: Zoloft on February 02, 2009, 01:08:28 AM

Title: Re: Calorimeter again!
Post by: Borek on February 02, 2009, 03:51:27 AM

Quote from: thesloc on February 02, 2009, 01:08:28 AM

But what is :delta: T for the calorimeter?

Initial temperature of calorimeter was identical to that of water it contained.

Title: Re: Calorimeter again!
Post by: Zoloft on February 02, 2009, 01:18:46 PM

Quote from: Borek on February 02, 2009, 03:51:27 AM

Quote from: thesloc on February 02, 2009, 01:08:28 AM

But what is :delta: T for the calorimeter?

Initial temperature of calorimeter was identical to that of water it contained.

But the heat capacity of the bomb calorimeter is 15.7 J/K without the water?  ???

So if the final temperature of bomb calorimeter is 32.4 and the initial is 23.6, then :delta: T is 8.8 K,

So using the heat capacity (C) of 15.7 J/K, I can solve for q using q = C :delta: T

So q is  138.16 J

No, the :delta: T of water is also 8.8, so I can easily get the q.

But the q of the sample? Since q = m x Cs x :delta: T,

would....the :delta: T be 32.4 - 98.7 ?

And then I could just write the equation as

Cs x 307g x -66.3 K = -(q_water + q_calorimeter)

And solve for Cs, right?

So Cs = [-(q_water + q_calorimeter)]/(307g x -66.3 K)

Title: Re: Calorimeter again!
Post by: Borek on February 02, 2009, 02:47:39 PM

No time to check details, but you seem to be on the right track.