Challenge Questions

When the rock is in the boat it displaces an amount of water equal to it's weight. When the rock is in the lake it displaces an amount of water equal to it's volume, unless it is less dense than the volume of water it displaces. If the rock is more dense than water then it will displace less water when it is thrown overboard. If the rock is somehow less dense than water (pumice? lava?) it will displace the same amount of water when thrown overboard as it did when on the boat. Therefore the level of the lake will decrease by:

[(150 lb water * 0.01604 cu.ft./lb) - (150 lb / density of rock)] / area of lake (sq.ft)

This formula works whether the rock is more dense than water or less dense than water.

Even in a very small lake, the change in level would be insignificant(virtually immeasureable) .

What if the rock is able to absorb water? Then the volume of water it displaces will be less and the lake's level will drop when the rock is submerged.

If the rock's density is less than the density of the water then it will float, leaving the lake level the same.

the porosity of the rock has no effect if the rock is in the boat, if the rock is in the water the pores of the rock will fill with water - this will happen very slowly so that the rock need's time to begin sinking. if the density of the rocks material without any pore is less then the density of water the rock will swim anyway and has no longtime effect to the level of the beautiful lake you are fishing on (maybe except of the beauty of the beautyful lake you are fishing on)!

Where do you buy neutronium sponge rocks that weight 150 pounds but do not displace appreciable water?

Please state alll assumptions.... ie .... is there any reaction between rock and water.. soluable?? is the rock pourous?? is temperature constant? is the lake constant volume? is there flows in or out of the lake? over what time period is the examination? is it steady state?

The rock, in the boat, is displacing 150 lbs of water. If the rock is thrown over board, the rock will still displace 150 lbs of water; however, the cubic feet of the mass of the rock comes into focus. If the rock is dense and occupies a small cubic area, then the level of the lake will be lowered by the cubic area displaced by the boat minus rock, divided by the cubic area of the boat plus the rock. The cubic volume of the lake will be divided into this result which will yield a miniscule percentage of increase in voume. The cube root squared of the volume will will yield the surface area increase.

If the rock is positively bouyant in water, then the lake water will only be displaced by the specific gravity of the rock per cubic area. The cubic volume of the lake will be divided into this result which will yield a miniscule percentage of increase in voume. The cube root squared of the volume will will yield the surface area increase.

a swimming boat in an lake of mercury/quicksilver would have a smaller effect to the 150 pound rock, the rock should swim after throwing him overboard in any case.

But where can you find a beutiful lake of mercury with fishes!

The boat and he rock are combined weight plus atmospheric pressure causing a pressure downwards resisted by surface tension pushing the water towards all the surrounding banks when the rock is tossed out of the boat the combined weight is lessend the boat assumes a new W/L displacement bobs up and causes less tension on the water surface allowing the water to recede .Archimedes a floating body displaces it's own weight in water when thrown out the rock resides on the bottom exerting no pressure on the surface of the lake. So the water level must return to its previous level being on the bottom of the lake the rocks displacement has little effect according to my own theorising. Could somebody bring in more factors regarding the rock on the bottom of thr lake

The water level remains the same. It doesn't matter if the rock is on top of the water or at the bottom of the water, it still displaces the same amount of water. So the water level doesn't change.

The level drops slightly. In the boat, the rock displaces 150 lbs of water. Once dropped overboard, the rock displaces the amount of water equal to its volume. Since the rock is more dense than water, the amount of water that the rock displaces inside the boat is more than its volume.

The level drops slightly. In the boat, the rock displaces 150 lbs of water. Once dropped overboard, the rock displaces the amount of water equal to its volume. Since the rock is more dense than water, the amount of water that the rock displaces inside the boat is more than its volume.

Plus the weight and volume of the boat

I initially thought this was too easy, since the boat will displace an amount of water equal to the weight of the rock. But the rock is undoubtedly heavier than water, and will sink to the bottom displacing a volume of water equivalent to the volume of the rock rather than the amount of water displaced by its weight while it is in the boat. The level of the lake should drop slightly.

the lake The displaced water has a weight equal to weight of the rock in the boat. The rock in the lake only displaces it's volume which is less than volume of diplaced water while the rack is in the boat. The level therfore rises.

What if the boat is a hovercraft? Then the lake would rise.

A hovercraft also displaces water equivalent to to its weight, thru the distributed air pressure under the skirt. The water level would still fall.

the noise of the hovercraft will chase away the fishes - the pressure of the hoovercraft is diaplacing the same amount of water of the hoovercrafts weight (including or excluding the rock).

The level of the lake will be changed by the difference the volume of 150 lb of water and the volume of the rock distributed over the difference between the surface area of the lake and the volume of the lake.

As rocks have a density greater than water, the 150-lb rock will, according to Archimedes principle, displace a volume of water with a weight of 150 lbs, i.e., greater than the volume of the rock. If the rock is thrown overboard, it only displaces an equal volume of water. So the level of the lake will actually LOWER when the rock is thrown overboard!

Actually I think my answer must be in the large list, but I have the necessity to post it. This question is relatively simple, but as have been discussed depend in the density of the rock, but most important is the volume of the lake. The question is about the level of the lake not to where the water level reach the boat.

So the change of adding a rock of 150 lb, almost 60Kg have no effect in the level of the lake besides the generation of a wave initiated in the point where the rock was throw. Now if we are talking about a small pool, then the result changes. But in general a lake is big enough to allow the addition of that body without an appreciable change in its level.

Here are capable persons to show us if it has not been done how the change depends in the total volume of the lake.

Thank WhoKnowsNothing

Assuming the rock is typical and has a greater density than water (i. e., not pumice or a thin-walled geode), the lake level will fall.

The amount of water displaced by the boat while the rock is in it is 150 lbs. greater than after the rock is tossed into the lake. When the rock leaves the boat, the boat rides higher in the water, so it is not pushing as much water upward around it. When the rock enters the water, the water level increases again but not by as much as when the rock was in the boat.

For example, if the density of the rock is 7 times that of water, its volume would be 0.34 cubic feet (the density of water is 62.4 lbs. / cubic foot). 150 lbs. of water has a volume of 2.4 cubic feet. The boat displaces an additional 2.4 cubic feet of water while it contains the rock, but the sunken rock only displaces 0.34 cubic feet.

Hello Challenger,

Why should you carry a rock in the boat unless it is ice (When you say "ON THE ROCKS" you mean pour the drink on the ice) to keep the fish from spoiling. I believe that your rock is ice.

150 lbs of ice will displace 68.1675 liters of water (Assuming that the specific gravity of water is 1.

Volume of 150 lbs of ice will be equal to 68.1675/0.919 = 74.17573449 liters.

When thrown into water, 0.919 portion of the ice which is 68.1675 liters of water will be displaced. Now since 150lbs is removed from the boat, the boat will be rise, as a result the level of the lake will remain the same.

Now when the ice melts, the volume above the water level will be added to the lake but molten ice under the water will result in a reduction of volume and the volume of water resulting from melting of the ice will be equal to 68.1675 liter.

In nut shell, the level of water in the lake will remain the same.

Job Thykkoottathil

if the rock is ice and the water is pure alcohol - you will need a therapy

Thanks for the kind remark.

Am I not to understand that the very nature of a boat in order to float it must displace water weight more that that of the vessel.That being true and if that boat has a rock that weighs say 100 lbs it must displace that weigh-volume also to maintain the same water line if not it sinks that volume now if you were to throw the rock over board , the boat will rise that fraction in the water (same volume less weight) and the lake would rise that fraction (the volume of the rock) that displaces the water. much like myself in the tub. of course throw a rock into the ocean you won't see any difference.

too much assumption here

(rock density has not been stated )

and what is " 150# " ??? ........ is that 150 pounds or 150 cubic centimetres ?

you must state units and if in doubt seek clarification. assume NOTHING.

but lets break it down into elements

when the rock is lifted , it should be lifted at 0.000000001 metres per second so as not to cause the boat to push down in the water which would result in a momentary rise in lake level.

when the rock leaves your hands and begins falling towards the water , the boat will rise and so too the lake level will begin to rise in a ( somewhat delayed ) momentary reactive until the rock hits the water and begins sinking towards equilibrium against the lake bottom

If the rock density is almost the same as that of water ( 0.99999999999 x )then there will be no change in lake level when the rock is dropped in.

If however the rock has a density of 3 kg per litre of space ( displacement ) versus water having a displacement of 1 kg per litre......

> then assuming 150# = 150 pounds

the lake will fall by 68 litres of displacement when the rock leaves your hands , but again rise by 1/3 of that ( 22.73 litres ) when the rock attains enlightened unification with the water ( translated as when it sinks in the water )

Therefore the lake rises by 45.45 litres relative to its volume.

howzat ?

do i get a prize ?

chocolate ?

gold stamp on my hand to show mummy ?

a dance with Claudia Shiffer ?

:)

.

ooops

typo in my post above

68 litres minus 22.73 litres = the lake _FALLS_ by 45.45 litres

.

pardon my fat fingers

.

( hope claudia likes fat fingers )

.

The water in the lake would fall. In the boat the rock displaces 150 lbs of water. In the water the rock only displaces water equal to its volume which is less because of its greater density. Instead of a rock think of a 150 lb chunk of lead

Nothing will happen to the water level of the lake itself because you are all ready on the lake with the rock but, when you throw it overboard the water covering the boat bottom surface will be lesser due to the change of weight pressure from the boat to the water.

nothing will happens to the level of lake

Nothing. any object immersed in a liquid is buoyed up by the liquid it displaces. the rock has affected the level by being in the boat already, by throwing it overboard, you have only got it out of your way.

I think the level will remain unchanged. It doesn't matter what the size of the rock. think of this, if the rock was at the back or the front of the boat it would apear that less of the boat is in the water but it would in fact still be displacing the same amount of water, because the boat would sink more on either end, thus when the rock is thrown overboard the displcement would be unchanged. It's allready stated that the rock weighs 150lbs so it doesn't matter what the rock's matter concist of.

Level goes down.

Archimedes Principle. The water level rises equal to the volume of the stone. However, the boat goes up by a volume of water equalling the weight of the stone. This rise being more than the other, overall the water level decreases.

bioramani

It's not just any old lake but a beautiful one. Probably you won't think the lake is so beautiful if it is raining, so it is a beautiful dry day. So the lake is evaporating.

Oh, and throwing the rock over the side of the boat will marginally reduce the level too, unless the brick floats, in which case it depends whether the water wets the rock or not. If it fully wets the floating rock the level will rise ever so slightly as the rock is sucked into the water. The net effect of partial wetting will depend on the difference between the wetting of the boat and the wetting of the rock (and on the total perimeters).

The level of the lake will go down. While the rock is in the boat it is displacing water of equal weight to the rock. Once the rock is thrown in the water the boat no longer displaces the water and the only water displaced is the volume taken up by the rock. Since the rock is denser than water the volume displaced is less so the lake level decreases.

The level of the lake would increase at least until I took my boat out of it because the fish are not going to bite after I throw the rock in....

The level of the lake goes down. When the rock is in your boat it is displacing its equivalent weight of water. When it is thrown out of the boat it only displaces its volume of water and the level of the lake goes down.

You are fishing on a beautiful lake. There is a 150 lb rock in your boat. It's in the way so you throw it overboard. What happens to the level of the lake?

Nothing. While the rock was in the boat it was causing the boat to displace the same amount of water as it would out of the boat and in the water. Only the location of the displacement would change.

It is not possible to answer this question accurately with the given information.

Water displacement is a function of volume and not weight.

If the rock were very dense and small, the boat would displace more water than the rock.

On the other hand, if the rock was not made of dense minerals, and the volume of the rock was relatively large, then the rock may displace more water when it is thown into the lake than it did while it was in the boat.

The size various of 150 lb. rocks may differ. A one cubic foot, 150 lb. rock would seem to displace one cubic foot of water. A two cubic foot, 150 lb. rock two cubic feet of water. This assumes there is no water absorbed into the rock and it does not float and any splash or wave effect is absent.

tommm

How about some other questions?

1. Is the boat on land or in the water?

2. Do you throw the rock onto land or in the water?

Easy Answer: If the boat is in the water and the rock ends up in the water, the lake level stays the same!

The level of the lake falls.

While in the boat it is displacing its weight of water - 150 pounds. When thrown overboard it only displaces its volume which is dependent on its specific gravity.

The above assumes that the rock has a SG greater than the SG of the lake water.

Ray Australia

A ripple forms and spreads radially form the point the rock enters the water, waves may hit the shore, no increase in water level.

The usual case would be that the rock is more dense than the water, so that the rock will sink to the bottom and the lake level will rise. This is because the rock displaces its relatively large water equivalent volume when in the boat, but can only displace its own smaller volume when out of the boat and resting on the bottom of the lake. Even if the lake is quite shallow and the rock does not fully sink it is still the case that the water level will fall. A special case is where the rock is less dense than the water - in this case it will float (if the water is deep enough to allow it) and there will be no change in the water level. If in this case the water is too shallow to allow the rock to float then, once again, the water level will fall.

ERRATUM - must have had brain fade when I responded. Obviously from the balance of the response I meant to say "down" in the second line. !!!!...note to self, take more care.

The Answer is down.

Because, as the rock sits on the boat it takes as much space in the lake as water would/lbs. And when you put the rock in the water it will take less room in the water because it weight more by volume than water does.

That is assuming you could actually measure it.

OK obviously you must be old to fish with rocks(OR be in some sort of Marfia gang and failed to mention Loose Lips Johnny was in the cement rock). So one assumes you just old, you suffer a heart attack throwing the rock into the lake, and you notice as you keel over into the lake, the water starting to lower, as you see your maker you inquire to why this happened, He replies your rock opened up a fissure on the lake bed and the water started to fill it, eventually you wake up on a life support machine and begin to write theories and questions for the mob at CR4, to which no one can actually answer because the information is wrong, PS The question forgot to mention the lake was also frozen over !!

Please MR CR4 stop writting idiotic questions, many people die in Bar room fight from knife fights from questions like this !!!

Nuffin'

the level will be same.

Nothing - the water displacement was already accounted for when it was in the boat.

i.e when a cargo boat is loaded, the boat sinks down to the level of the loading line. In reality the water level as risen by a micro-fracational percentage.

It now does not matter whether the cargo is on the boat or thrown overboard - the water level will remain the same.

The average level of the lake will not change, because your boat will rise in the same time and for the same volume as the rock will enter the water (Archimede law). If fact, the level of the lack changed when you loaded your boat at the wharf.

Besides,

1- during the dropping time, the boat will rise and the rock will not be in the water yet, so the level of the lack will fall before rising back to the initial level.

2- The impact of the rock entering the water and the movements of the boat will generate waves propagating all around the lake. So the ponctual levels of the lake will change until fading.

Those two last points seem negligeable in a first appraoch, but think about the level of a whole lake compared to the volume of a 150 lb rock... The waves will create much more variantions.

Greg

Ok, I'm standing in a boat with a 150lbs rock in my hands and I am throwing it in the lake. First thing that's going to happen is when I throw the rock, the lake level will go down for mabe a fraction of a second "while the rock is in they air" as the boat goes up, then it would go back to it,s original level after the rock is totally submerged". But what about the ripples? Up, down, up, down...? I guess it's all in your perspective. What about the density and size of the rock? More density less room = water down.

Tit for tat,Weight by volume of the rock compared to the water weight by volume. would tell you the final answer "Assuming the rock is not porous".

if you was to add a drop of water to the ocean, did it make a difference? Yes it did, there is now one more drop in that ocean. Not like it actually affected a whole lot of things, but it made a difference.

Infinitely big and inffinitly small... time and mass

B.T.W. My wife thinks we're all a bunch of nuts in this forum… LMAO!!!

rock in the boat displaces the water equal to its weight. in terms of volume more amount of water is displaced than the volume of the rock, hence the level of water raises. when the rock is thrown overboard the level dips.

goes down Archimedes if boat was floating with rock, then displacment relative to density of rock, as rock is heaver than water when rock placed in lake the displacment was volume of rock only...

the level remains the same- the amount of water displaced by the boat with the rock in it is reduced by the amount of water the rock displaces when thrown overboard. it equalizes...

Andrew Holmbeck

Did we throw the rocks in the water? Overboard could have ended up on the shore, to which the level of the lake would go down due to the weight of the boat being less.

There you go.. the rock could also have landed in another boat. Been on lake Powel during the july 4th holiday could happen.

remember the question:

you are sitting in the boat on the lake not they are sitting in boats.....

Nothing! The weight of the rock has already dispaced the water via the boat. When you throw it overboard, the boat rises but the water level in the lake remains ther same since the boat has risen.

but the rock has less volume than the hull of the boat. unless it has the same density as the liquid it can not displace the same volume that the 150 lbs displace via the boat.

That would be true if you had a boat with 150 boat and you thru it overboat now the lake remains at the same level.

Come on Guys

Mental image - cross section through boat and rock.(basket ballsize).

1.) boat with rock on board

2.) boat with rock over the water ( rock the size of basket ball)

Unless the rock floats it will displace more water via the boats hull than by displacement of its own size in the water.

I believe it would remain the same as the net weight added to the lake remains the same. When the 150lb rock is thrown into the lake, the weight in the boat reduces by the same amount added to the lake.

It is important to remember that once the rock sinks (most do) its weight is irrelevant. The only thing that matters in regard to the lake's level once the rock is submerged is the displacement of the rock. In the boat the rock was displacing approximately 2.4 cubic feet of water. Submerged, the rock (lets assume that the rock is a diamond, cause they're nice rocks) would only displace approximately .68 cubic feet of water. This 1.72 cubic feet decrease in displacement of the water would result in the level of the lake lowering. With over 250 square miles of water within 2000 miles of shoreline you probably wouldn't notice it much here, though.

The remainder of the rock's weight twists the bottom of the lake out of shape, microscopically lowering the lake level more, so it is not irrelevant, but it is inconsequential.

The MUD on the ground of the lake will be kompressed - and separates in this way the water from the ground material; there cannot be any effect to the lake level (its a filled porose matter which is separated to its origins - water and ground-material)

You are fishing on a beautiful lake. There is a 150 lb rock in your boat. It's in the way so you throw it overboard. What happens to the level of the lake? The lake level should remain the same, as the lake level was displaced when you launched your boat

Wrong! See post 49.

I think we have covered this subject enough.

Why would a fisherman need 150 lbs. of balast in a small fishing boat? Couldn't he/she just bring a 150 lb. friend to keep warm with instead?

The level of the lake would rise proportionatly to the surface area of the rock. Small rock, small rise, large rock, larger rise.

Nothing.

The water displaced by throwing the rock in the lake was already displaced by having it in the boat (which sits below water level). Therefore, the rock may move, but the water level staysd the same.

There will be no change. The rock has already displaced it's mass while it was in the boat. So, the boat will ride higher, but the lake level will not change.

At first you might say nothing happens to the level of the level of the lake, because both the boat and the rock are still in the lake but you would be incorrect. imagine if you will that there is a 150 lb rock with a volume of 1 cubic foot inside a plastic airtight sphere with an volume of 4 cubic feet. if you were to place this rock inside a sphere in a lake it would most certainly displace more than 1 cubic foot of water. now if you remove the rock from the sphere and drop in into the lake (displacing 1 foot of water) you would have a very light sphere left which would not displace much water at all. this is the same as what is happenign with the boat/rock. removing the rock from the boat lowers the level of the lake

I think this is the most accurate answer but with a complicated example , the displacement of water while in the boat will be the rock weight (in a form of sinking volume of water equivalent to the rock weight the boat will do) but when it is in the lake it will sink so it will only displace it's volume. and because the density of the rock in general is higher than water, then water level in the lake will be less.

The lake becomes lower. In the boat, the rock displaces its mass of water. thrown overboard, with rock sinking (rock density greater than water) it displaces only its volume of water.

Hwgthrice

There would be effectivly no change in the lakes level. The boat displaces water equal in mass to the boat and its contents elevating the lakes level just the same as slipping into a saturday night bath. Once the rock is removed and placed into the very same lake the boat is positioned on, the displacement of water by the rock will shift from the boat to the lake bottom, assuming a ten ton guppy doesnt swallow and swim off with the rock as it heads for the bottom of the lake. The displacement of the lake or lake level would still be unchanged.

The real question is:

When the rock was thrown overboard, where did it land? (and who was tied to it)?

How did the crow drink from the glass that was half full, dropped stones in untill he raised the level to suit his needs. does the same not apply to the one who eats the crow

The level of the lake goes down. Because the rock displaces more but weight than it does by volume.

Augie Palmer

Oxford Science

On my mind!

If rock has a relative density less than or equal to that of water water the level of the lake remains the same. If the rock has a relative density greater than that of water the volume of water equal to its weight will no longer be being displaced when it is removed from the boat and the level of the lake will go down.

I'd like to think this is a relatively simple question. You're in the boat, so the boat is already displacing the water of the lake making it higher than normal. Having the rock in the boat increases displacement, and therefore water level. By taking that rock and throwing it overboard the boat is displacing less, which means the water level is lower relative to the boat, but the rock is being thrown in the water, so that is the displacement that your boat just lost. So to answer, nothing happens to the water level. Whether the rock is in the boat or in the lake, it's still displacing the same amount of water. The only way you could change the water level is to take your boat out of the water.
Dan, Electrical Engineering Student

Nothing, the rock is still displacing 150lbs of water

Been away (busy) for a while. So someone has summoned Archimedes from the depths? To what extent will this drop (like the level in the lake)? This could resurrect streaking in public, and cause a depletion of the nations sunscreen reserves.

Nothing measurable, I threw it on to the shore and the lake is very large.

Nothing happens to the level of the lake. The rock in the boat already displaces it's density. If the rock were less dense than water, then the rock would float, but still not affect the level of the lake - displacement total would not change.

While the rock is in the boat it will displace more water, the rock is able to displace more water in the boat because of its weight combined with the greater surface area on the bottom on the boat. Once in the water the rock it will only displace its volume.

thank you, thank you

Brett

Nothing. The rock was in the boat and should have been already displacing water so if trown over board it should still displace the same amount of water as it would inside the boat. Also was it thrown over board on land or in the water? Finally if it did displace water 150 lbs. on a lake would not displace a measureable amount of water.

Presume the rock density is greater than water (no floating), what ever rock weighs causes boat to displace xtra volume of water equal to the weight of the rock which caused the lake level to increase when boat & rock placed in water. When rock thrown overboard, boat no longer displaces water equal to weight of the rock, so lake level goes down.

But keep in mind, the rock has volume which incrementally raises the lake level when rock is in the water, but incremental increase is less than the decrease when thrown out of boat into water. Ans: Lake level goes down.

If rock had same density as water, lake level would not change upon getting deep sixed.

If rock were a 150 Lbm 'almost' black hole (infinite density, V=0), lake level would go down as though (150/62.4) cu-ft of water had been removed from the lake. dactheco

PS: I know, almost-BH-rock would sink in turf below lake, no matter, since volume of rock is ~zero. Too wordy...

Maybe someone has already said this,

Assuming the rock is non-porous and we don't get picky about how hot a day it is, water temperature, the rock's temperature, how much evaporation is occurring by a misting effect when the rock impacts the water, how hard is the wind blowing, surface tension altered by the purity, how many under-inflated inner tubes are in the lake or some other ridiculous variant. I think this is how I think it works.

On board, the rock will displace the same volume as 150lbs of that particular lake water would displace.

If the rock's density greater than the density of water the lake will go down.

If the rock's density is equal to the density of that lake water there would be no change in the level of the lake.

If the rock's density is less than that of water it would make the water rise.

In reality, in a "lake", I'm thinking of a true lake, where boats can move about, you'd probably would find it hard to measure much less see a visual change.

If the rock's density is less than that of water it would make the water rise.

The level of the lake will be unchanged!

Hmm, I would like to know more of the physical size of the rock, that is, a rock containing heavy elements such Galena etc, & has a physical size of 1/4 Mtr square will have the same weight as a Limestone rock of the same weight but the Limestone rock will be 1Mtr square, hence displacing more water once lifted from the boat & placed into the water. The boats displacement will be less of course & it will rise, BUT, the rocks physical size once in the water will have a different affect on the lakes level. The level I feel will rise if the Rock is of a physical substantial size or it will lower is the physical size is less than that of the boats changed displacement.... phew... sorry to confuse anyone... I don't know the mathematical formulea for this but is someone does.. please let me know... :)

cheers for now

LOWIQJO.. p/s.. some great answers in here, a good site..

Displacement is displacement. It matters not how it is applied. The level won't change.

If the rock is of greater density than water, the level of the lake will decrease slightly, as the full weight equivalent of water is no longer displaced. This is supposing that such a decrease doesn't cause a breakthrough from a very close and higher body of water. If the rock floats the level will be unaffected since the full weight of the rock will still be displaced.

Gentle Miant

Not sure that I agree with your conclusion. I think that it would only be so for a rock density exactly equal to that of water, and that, if the rock floats, the lake will rise by the differential volume displaced.

Assumptions:

1) the boat is floating on the lake (it is not resting on the bottom or supported by tension from a rope to a pier (etc)

2) the lake is not bound by a wier or dam wall that restricts any increase in height of the water, nor will an increase in height weaken any structure on the edge causing its collapse and leakage of water (or indeed ingress of more water)

3) loss of water by evaporation, splashing, or absorption into the rock will not occur.

4) there will be no chemical reaction causing the rock to dissolve or react with the water (or anything in it)

5) the rock has a density > 1 and sinks completely, i.e. i it sinks so is completely submerged.

6) The rock is thrown overboard into the lake, not onto a pier, or onto the shore.

7) The "level" of the lake is the mean height difference between some fixed datum and the surface (i.e. the air-water interface) of the lake. This specifically DOES NOT include the surface of the lake that is under the boat.

8) gain of water by rainfall, runoff, or any other method is excluded.

My proposed solution: As the rock is lifted (more specifically as it accelerates with a component of that acceleration being against gravity), the force acting on the bottom of the boat increases, pushing the boat deeper into the water thus increasing the level of the lake. As the rock is released, but before the rock impacts the water, the level of the lake lowers to a level below the initial height. As the rock enters the water the mean level increases as the rock enters the water.

[If the rock had a density < 1, the water level may increase slightly above the initial level as the rock submerged slightly below its equilibrium level, before the slight overshoot and subsequent ringing dies down. The lake would return to the original level.]

[If the rock had a density = 1, thee would be no "ringing", and the end result would be as in the d < 1 example above]

If the rock has a density > 1 (see assumption 5) it was originally displacing it's weight in water. As it enters the water and sinks it now displaces it's VOLUME of water. With a d > 1 the volume is less that the volume of an equivalent mass of water, thus the level raises from the minimum, but not quite to the level it was when the rock was in the boat.

So, with the assumptions above, the level first rises (as the rock is accelerated upwards) then falls (as the rock is released) then rises again (as the rock enters the water). The final water level is slightly lower than the initial level.

sublimation is just an effect to the surface and includes all ingredients of the frozen liquid!