Asked by justin
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
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
Answers
Answered by
DrBob222
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.
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.
Answered by
justin
thanks!!!! it was an exercise from study guide it just gave the answer not the solution. might be correction
There are no AI answers yet. The ability to request AI answers is coming soon!