## Aquarium Problem #3

Posted by DP on April 17, 2013 – 10:33 am

A 2 foot by 3 foot aquarium is filled with 9 inches of water. A 1 foot ice cube is placed in the corner. How much will the water rise once the ice melts?

Posted by DP on April 17, 2013 – 10:33 am

April 17th, 2013 at 2:49 pm

Well, ice is lighter than water, so 1 cu ft of ice would melt to a bit less than 1 cu ft of water. But I suppose we disregard this here.

So, with 4.5 cu ft of water already in the tank, another 1 cu ft takes this to 5.5.

This covers 6 sq ft, so the depth is now 11 inches, a rise of 2 inches.

Good to see someone contributing some puzzles to ToM after such a long break. Where’s Chris these days? Let’s hope these Aquarium ones break the ice (ha,ha!) and we now get more on a regular basis.

April 17th, 2013 at 8:03 pm

Hi Wiz. You got the first two aquarium problems right.

In this case, there is an initial rise when you put the ice in. What’s wanted is the rise after that – i.e. the rise entirely due to the ice melting.

April 18th, 2013 at 12:45 am

Ah Chris, it’s good to see you’re still around to pick holes in my reasoning as you always have done!

Just wondering if there might be a slight inconsistency here. The first two problems ask for the water level rise AS A RESULT OF putting the metal cubes in. Here you’re saying we want the water level rise, NOT as a result of putting the ice cube in but as a result of it melting.

As always, the difference seems to come down to how you interpret the question.

April 18th, 2013 at 6:59 am

Hi Wiz. It’s very good to see you back here too.

I agree, the question hasn’t taken great pains to be clear on the point, especially in view of the previous two problems. I’m completely confident that I’ve interpreted the question correctly.

Here’s a big (but oblique) clue. Assuming a relative density for water of exactly 1.0 and for ice of exactly 0.92, you’d have got an initial rise of exactly 1.84″.

Much as I love picking holes in your reasoning, you often don’t leave me holes to pick. You also often give very nice reasoning (even when it’s wrong ). On this occasion I’m not picking holes in your reasoning as there aren’t any.

April 18th, 2013 at 11:00 am

You both make some very valid points. Here on ToM it would not be uncommon to list your solution, including any assumptions you made along the way that were not explicit in the problem statement.

I would say that a good way to approach this one might be to analyze each part and show the results. Something like “When the ice is initially placed in the tank, the water level rises from 9 inches to X. When the ice melts, the water level is now at Z (including calcs), which is Y inches above X.”

I was hoping the “ice is less dense than water” would come up. I was also curious about this, so I did some quick research of my own. I always “knew” that was “true”, but never know the properties/values. As Chris pointed out, 0.92 is the magic number. Stated in a different way, water gains approx. 8% more volume when it becomes frozen. That seems to be a significant enough value to consider here. More than I initially thought anyway.

I was hoping this might trip someone up, maybe thinking that the ice above the water level would be all that contributes to the overall water level rise. In actaullity the Ice below will effectively drop the water level on it’s own when it melts.

I guess we should also assume that the ice is not floating just above the tank’s bottom surface. I’ll address this in the following post, but maybe someone could check my math.

This brings up a thought for Aquarium Problem #4, coming soon to a ToM site near you.

April 18th, 2013 at 11:17 am

As for the floating cube…

Referencing Prob #1, the initial water level (with the cube, initially assuming it does not float) is 10.8 inches. The ice displaces 10.8×12x12 = 1555.2 cubic inches = 0.9 cubic feet of water. The weight is equivalent to 0.9/0.92 = 0.97826 [approx.] cubic feet of ice. The ice is 1 cubic foot, so it will not float, and 0.02174 [approx.] feet (0.26 inches) will appear above the water’s surface.

April 18th, 2013 at 1:41 pm

…oops. I just realised that I goofed. Despite my calculation, I hadn’t noticed that the ice block wasn’t floating.

I note that DP has just pointed that out.

Shame, that means that I really have to assume a density for ice e.g. 0.92. Still, the question is easy to do.

April 18th, 2013 at 8:46 pm

Now I realise that the ice wasn’t floating, I agree that the initial rise is 1.8 inches – exactly as for Aquarium Problem #1.

I also wish I’d suggested using 0.9 as the density of ice. Then the ice block (if floating) would displace 0.9 cu ft of water and that would give a rise of 0.9/6 ft = 1.8 inches. As this is the same rise as in problem 1, we’d then have that the ice would be exactly at the threshold of floating/not floating, and then, when the ice melts there would be no further change in the water level.

Assuming ice has relative density 0.92, then it is definitely (but only just) not floating. So the initial rise is still 1.8 inches. The cu ft of ice melts to 0.92 cu ft of water, so the final depth is (4.50 + 0.92)/6 ft = 10.84 inches. i.e. the additional rise due to melting alone is 1.84 – 1.8 = 0.04 inches.

May 6th, 2013 at 8:18 am

I’ve just realized that I never gave an official solution. Chris has it in the final paragraph of post #8.

It looks like these aquarium problems were successful in ending the ToM drought. Keep ‘em coming!