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.
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
3 answers
47c
no