Calorimetry problem [closed]

Question

The second picture shows the textbook solution which I don't find convincing and the third picture shows my work.

I don't understand why will the block fall 40 cm toward the bottom of the vessel as given in the solution. According to me, buoyant force will act upon the block and it should fall to a height lesser than 40cm. If that is so, then we cannot write

$mgh = loss in energy$ The loss in energy will be less.

Where am I wrong? Why did the solution did not take buoyant force into consideration?

Also, when the spring is attached to the block, the block has $0.5 kx^2$ elastic potential energy. After the support of the spring has been broken, the block no longer possesses this energy.

Where does this elastic potential energy go?

Answer 1

It will fall to the bottom if the density of the block is higher than that of the water. The potential energy of the spring is dissipated by friction between the coils of the spring and by inelastic collision of the coils when the spring recoils. Your real question should be "what is the mechanism by which the gravitational potential energy of the block is converted into an internal energy rise for both the block and the water?"

Answer 2

The block will indeed fall to the bottom of the container, as long as the density of the block is greater than that of the water.

But the solution supplied in the original post neglects the fact that there is an equal volume of water that is raised up to the occupy the former position of the block.

The water that is raised up has the same volume as the block, and therefore $\frac16$ the mass, or $0.2$ kg

So the correct expression for the energy released by the block in falling should be:$$Energy =(1.2\times 9.8 \times 0.40)-(0.2\times 9.8 \times 0.40)$$

The same result would be obtained if we considered the work done by the net force (gravity minus buoyancy) as it pulls the block to the bottom.

So the correct energy release would be $3.92$ joules, not $4.704$ joules ($4.707$ is a typo)

With regard to the stretched spring, the net force exerted by the block on the submerged spring is $9.8$ newtons, so the extension $x$ is given by $$x = \frac{F_{Net}}{k}=\frac{9.8}{200}=0.049 \text{ meters}$$Putting this value into the formula for the stored energy, we get:$$E_{Spring}=0.5 \times 200 \times 0.049^2=0.2401\text{ Joules}$$This energy is released when the block starts to fall. After the spring stops thrashing around, this energy is also converted into heat.

So the total energy to be distributed as a temperature increase to the block plus water is $4.16$ joules.