(1) Heat absorbed by water:
Q absorption=cm△t=4.2×103J/(kg?℃)×100kg×20℃=8.4×106J;
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(2)G total=m total g=ρV total g=ρ(VA+VB)g=1.8×103kg/m3×(5×10-4m3+5×10-4m3)×10N/kg =18N,
(3) Under normal circumstances, when the water level is the highest, the tension on the rope is the smallest. At this time,
F total float =G total-F=18N-10N= 8N
FB float=ρwater gVB=1.0×103kg/m3×5×10-4m3×10N/kg=5N
FA float=F total float-FB float=8N -5N=3N
Answer: (1) The heat absorbed by water is 8.4×106J;
(2) The volumes of A and B are both 500cm3, and the total gravity of A and B is 18N;
(3) Under normal circumstances, when the water level in the water tank is the highest, the buoyancy force on A is 3N.