REST OF THE PROBLEM...
20.0-kg gold bar at 25.0C is placed in large, insulated 0.800-kg glass container at 15.0C and 2.00 kg of water 25.0C. Calculated and final equilibrium temperature.
The book have an answer of 26.0C
PLEASE HELP ME I HAVE DONE THIS PROBLEM THREE TIME AND STILL DON'T UNDERSTAND HOW THIS ANSWER CAME ABOUT, PLEASE SHOW ME THANK YOU
2007-11-23 14:06:27 · 3 answers · asked by markeesha h 1 in Science & Mathematics ➔ Physics
No way it's 26.0C. The book is wrong.
The final answer will be around 24.8C - 24.9C. I would be willing to put down $10 that says it's in that range.
Edit:
Qgained by glass = Qlost by water + Qlost by gold
Q = mc(DeltaT)
800g(0.84 J/(g.C)(Teq - 15C) = 2,000g(4.184J/g.C)(25C - Teq) + 20,000g(0.129J/g.C)(25C - Teq)
672J/C(Teq) - 10,080J = 209,200J - 8368J/C(Teq) + 64,500J - 2580(Teq)
11,620J/C(Teq) = 283,780J
Teq = (283,780J / 11,620J)/C
>>>((( Teq = 24.42C )))<<<
I lost $10.
2007-11-23 14:24:02 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Live in the real world. Who puts 20 kg of gold into a container and then worries about the temperature?????!!!!!
2007-11-24 06:54:08 · answer #2 · answered by za 7 · 0⤊ 0⤋
You do not have the specific heat density of the gold.... it is normally in joules*kg/kelvin or similar format. It should be in your index or question summary.
2007-11-23 22:11:15 · answer #3 · answered by blue_zoo22 3 · 0⤊ 0⤋