Asked by kelly
what is the final temperature when a 3.0 kg gold bar at 99 degrees is dropped into a .22 kg of water at 25 degrees
Answers
Answered by
drwls
Heat will flow from the gold to the water until an equilbrium temperature T is reached. You will need the specific heats of gold (C1) and water (C2 = 1.00 kCal/kg C).
M2*C2*(99 - T) = M1*C1*(T - 25)
(99-T)/(T-25) = M1*C1/(M2*C2)
Solve for T (the final temperature). m1 is the mass of gold and M2 is the mass of water.
M2*C2*(99 - T) = M1*C1*(T - 25)
(99-T)/(T-25) = M1*C1/(M2*C2)
Solve for T (the final temperature). m1 is the mass of gold and M2 is the mass of water.
Answered by
sohab
47c
Answered by
yara
no
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