Posted by **lyne** on Friday, January 2, 2009 at 7:52pm.

A 46.2 gram sample of copper is heated to 95.4 degrees celcius and then place into a calorimeter containing 75.0 grams of water at 19.6 degrees celcius. The final temp of the water and copper is 21.8 degrees celcius. Calculate the specific heat capacity of copper, assuming all heat lost by the copper is gained by the water.

Answer in J/gram degree celcius

how do i set this problem up? I already sorted the information.

- chemistry HELP -
**DrBob222**, Friday, January 2, 2009 at 8:37pm
[mass Cu x specific heat Cu x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0

Solve for specific heat Cu. It's easier to put in all the numbers first into the above, do the math, then solve for specific heat Cu last. Post your work if you get stuck. For your information, note the correct spelling of Celsius.

- chemistry HELP -
**GK**, Friday, January 2, 2009 at 8:41pm
heat = (sp. heat)(grams)(temp. change)

or q = (C)(m)(Tf-Ti)

Heat released by Cu = Heat absorbed by H2O

(C)(46.2g)(95.4C-21.8C)= (4.18J/g.C)(75.0g)(21.8C-19.6C)

Solve for C to get the specific heat of copper.

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