Water and ice, scales [closed]

Question

There are scales with an identical bowl on each pan. We pour some amount of water into the first bowl. We pour less water and put an ice cube into the second bowl so the level of water in both bowls is identical. Will the scales be in balance

(a) in the beginning?

(b) when the ice has melted?

Assume water (in the liquid state) has constant density.

My reasoning:
(a) in the beginning, the ice cube's weight will be cancelled out with the buoyancy force from the water and it's obvious that the weight of the water in bowl one is greater than the water in bowl two, so the scales will not be in balance
(b) when the ice melts, the level of water in both bowls will be the same (because the ice will turn into as much water as it lifted in the beginning), so the scales will be in balance.

The model answer to the exercise is that the scales will always remain in balance and I can't find a mistake in my reasoning. Thanks in advance for your help.

Answer 1

The error in your reasoning in (a) is that since the water exerts a buoyancy force on the ice that is equal to the ice's weight, so by Newton's Third Law the ice must exert an equal and opposite force on the water. So the total downward force on the water is its own weight plus the weight of the ice. Once you follow this through by calculating the force on the bowl, then the force on the scales, you find that the downward force on the scales will be the weight of the ice plus the weight of the water plus the weight of the bowl.

The question is therefore which weighs more - the ice or the water displaced by the ice (which is the difference between the amounts of water on the two sides of the scales). And Archimedes' principle tells us that these weigh the same. So the will balance in (a).

You could also argue that since the scales will balance in (b), and since melting the ice does not change it weight, then the scales will also balance in (a).