Posted by **justin** on Tuesday, August 20, 2013 at 12:48pm.

hi this seems simple but i cant figure it out.

If a 0.46 kg hot copper pan lost 9.8x103

J of heat as it cooled down to 54°C, what was its

original temperature?

I used the 24.44 j/molK for heat capacity, is not this one? found on wikipedia: copper

this is how i derive my equation...

-9.8x10^3=7.24mol(24.44j/molK)(327-x)

the answer is 77C but im getting 109.39C :( please help thanks

- Chemistry -
**DrBob222**, Tuesday, August 20, 2013 at 3:43pm
I used 0.385 J/g*C and came up with the same answer. You can prove that 109.39 is correct (although that's too many significant figures to use) this way.

109-54 = 55.39

55.39 x 0.385 J/g x 460 g = 980.9 J.

55.39 x 24.44 x (460/63.54) = 980.03 J.

- Chemistry -
**justin**, Tuesday, August 20, 2013 at 4:32pm
thanks!!!! it was an exercise from study guide it just gave the answer not the solution. might be correction

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