Theoretical swimming pool question I need help with

[GeorgieBoy]

If a 20,000 litre swimming pool has a water temperature of 20 degrees centigrade, and you remove 500l at a time and heat the water to 90c then return it to the pool, how many litres do you need to heat and return until the pools temperature is at 30c? And how many for 35c?

Thanks for any help

 

[Deleted member 4993]

If a 20,000 litre swimming pool has a water temperature of 20 degrees centigrade, and you remove 500l at a time and heat the water to 90c then return it to the pool, how many litres do you need to heat and return until the pools temperature is at 30c? And how many for 35c?

Thanks for any help

How many 500 L make 20000 L?

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem

 

[Dr.Peterson]

If a 20,000 litre swimming pool has a water temperature of 20 degrees centigrade, and you remove 500l at a time and heat the water to 90c then return it to the pool, how many litres do you need to heat and return until the pools temperature is at 30c? And how many for 35c?

Thanks for any help

You say this is theoretical; does that mean it's for a class? It sounds well beyond arithmetic; and it also requires some physics, I'd think. Can you tell us about the context of the problem, including why you are asking, and how much you know?

 

[GeorgieBoy]

You say this is theoretical; does that mean it's for a class? It sounds well beyond arithmetic; and it also requires some physics, I'd think. Can you tell us about the context of the problem, including why you are asking, and how much you know?

It’s not for class, sorry if it’s the wrong place to post. I am working on the feasibility of an invention which is a portable solar powered pool heater. It probably is more Physics than Maths but Physicists are harder to find and thought someone here might be able to work it out

 

[GeorgieBoy]

How many 500 L make 20000 L?

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem

I haven’t tried too much yet, I guess an easy place to start is if you heated half the pool to 90c then the average temperature would become half way between what it started and and what it was heated too which would be 55c.

It gets a bit complicated for me going on from there

 

[Deleted member 4993]

I haven’t tried too much yet, I guess an easy place to start is if you heated half the pool to 90c then the average temperature would become half way between what it started and and what it was heated too which would be 55c.

It gets a bit complicated for me going on from there

How much physics of heat content do you know? What level of mathematics do you know?

If you mix 10 L of water at 20 C with 2 L of water at 90 C - what would be the temperature of the resulting mixture?

 

[Steven G]

Suppose that the water is at room temperature at 20C. Also assume that the room is always at 20C. You remove some water, heat it to 90C and return it to the water. Now if you do this once per day, the water temperature will return to 20C. The answer that you want is a function of time. That is if the time between your putting in a container of hot water is great, then the answer is that you'll never have the water a30C
Please think what I am saying.

 

[Mr. Bland]

I interpret this scenario to be a volume of water perfectly insulated from all external temperature change that does not radiate any of its thermal energy. The removed water is uniformly heated to 90 degrees, which then perfectly distributes to the entire pool once returned.

Answering this question requires a unit of energy to serve as a medium for the computation. Let us call this unit the "liter-degree", which represents 1 degree Celsius per liter of water. (It's conceivable that there's a name for this already, but I don't know what it is off the top of my head. (Kilo-)Calorie comes close, but that's defined per kilogram). The pool starts with some number of liter-degrees, and it will end with some higher number of liter-degrees. Each time some of the water is removed, heated and returned, the net effect is that the full volume of water receives some increase in total number of liter-degrees, thereby uniformly raising its temperature.

The energy contents in liter-degrees of some volume of water will be the amount of water in liters multiplied by its temperature in degrees Celsius.

• How many liter-degrees does the pool start with, and how many will the pool have once fully heated to the target temperature?
• How many liter-degrees are added to the pool each time a portion of the water is removed, heated, then returned?

Once you know by how much the temperature increases each remove-heat-return cycle and how much the temperature needs to increase total, it will be trivial to determine how many times the remove-heat-return cycle is needed.
__________

EDIT:
It occurs to me that this isn't a simple division problem, because the temperature of the water being removed each time is slightly higher than it was the previous time. The number of liter-degrees being added each cycle is therefore less each time you do it (the water you remove keeps getting closer to 90 degrees each time you take out 500 L).

This is ostensibly a calculus question, but with discrete intervals. I regrettably don't have a lot of experience in this area, so I don't know that I could put together a formula without simulating it.

 

[GeorgieBoy]

I interpret this scenario to be a volume of water perfectly insulated from all external temperature change that does not radiate any of its thermal energy. The removed water is uniformly heated to 90 degrees, which then perfectly distributes to the entire pool once returned.

Answering this question requires a unit of energy to serve as a medium for the computation. Let us call this unit the "liter-degree", which represents 1 degree Celsius per liter of water. (It's conceivable that there's a name for this already, but I don't know what it is off the top of my head. (Kilo-)Calorie comes close, but that's defined per kilogram). The pool starts with some number of liter-degrees, and it will end with some higher number of liter-degrees. Each time some of the water is removed, heated and returned, the net effect is that the full volume of water receives some increase in total number of liter-degrees, thereby uniformly raising its temperature.

The energy contents in liter-degrees of some volume of water will be the amount of water in liters multiplied by its temperature in degrees Celsius.

• How many liter-degrees does the pool start with, and how many will the pool have once fully heated to the target temperature?
• How many liter-degrees are added to the pool each time a portion of the water is removed, heated, then returned?

Once you know by how much the temperature increases each remove-heat-return cycle and how much the temperature needs to increase total, it will be trivial to determine how many times the remove-heat-return cycle is needed.
__________

EDIT:
It occurs to me that this isn't a simple division problem, because the temperature of the water being removed each time is slightly higher than it was the previous time. The number of liter-degrees being added each cycle is therefore less each time you do it (the water you remove keeps getting closer to 90 degrees each time you take out 500 L).

This is ostensibly a calculus question, but with discrete intervals. I regrettably don't have a lot of experience in this area, so I don't know that I could put together a formula without simulating it.

Thanks for your excellent reply.

Maybe someone could just answer me this. I presumed if you removed 50% of the water (10,000 l), heated to 90c from 20c and returned it the temperature would be half way between 20c and 90c so 55c.
Therefore if you just removed 25% of the water (5000 l) and returned it the temperature would be half way between 20c and 55c so 37.5, does that make sense or does it not calculate like that?

 

[Deleted member 4993]

If a 20,000 litre swimming pool has a water temperature of 20 degrees centigrade, and you remove 500l at a time and heat the water to 90c then return it to the pool, how many litres do you need to heat and return until the pools temperature is at 30c? And how many for 35c?

Thanks for any help

Let us assume that the energy required to heat-up 1000 litre (=1 kL)of fluid by 1 C is H (calorie) and the energy released during cooling down 1kL fluid by 1 C is H.

Let us assume that the Volume of fluid removed (in total) is V (kL). If we assume that the

energy released by hot fluid (90 → 30) with a Volume of V kL = energy absorbed by the cooler fluid (20 → 30) with a volume (20 - V) kL

You can calculate V (kL) from the relationship stated above.

Think about the assumptions we need to make for the process to complete without complications.

 

[The Highlander]

Maybe someone could just answer me this. I presumed if you removed 50% of the water (10,000 l), heated to 90c from 20c and returned it the temperature would be half way between 20c and 90c so 55c.
Therefore if you just removed 25% of the water (5000 l) and returned it the temperature would be half way between 20c and 55c so 37.5, does that make sense or does it not calculate like that?

Assuming that 1 litre of water has a mass of 1kg, then, if you add together two volumes of
water (\(\displaystyle V_{1} \& V_{2}\) litres) with respective temperatures (\(\displaystyle T_{1} \& T_{2}\) °C), the resulting temperature
(\(\displaystyle T_{f}\) °C) of the mix will be:-

\(\displaystyle T_{f}= \frac{V_{1}T_{1}+V_{2}T_{2}}{V_{1}+V_{2}}\)​

So heating half the water to 90 °C and returning it to the pool should produce a final temperature of: 55 °C whilst if you just heated 5 kl to 90 °C and added that back to the 15 kl sitting at 20 °C then you should, indeed, get a final temperature of 37.5 °C.

Your 'assumptions' are, therefore, theoretically correct but, dealing with such large volumes of water, if you just dump the heated water into the pool you are more likely to get a temperature gradient unless you have some way of mixing the water to produces a homogenous result.

Furthermore, the actual formula for determining the resultant temperature is based on mass not volume, therefore, some 'adjustments' would likely be necessary to account for the fact that when heated to 90 °C the assumption that 1 litre of water has a mass of 1kg no longer holds! (Though you could probably ignore that since it's just the volume, not the mass, of the heated water that has changed; there might be an increase of 3 to 4% in the volume of the heated water so heating 500l to 90 °C might "produce" an "extra" 15-20 litres of water whilst heating 5 kl could increase the volume by 150-200 litres!).

 

[Deleted member 4993]

If a 20,000 litre swimming pool has a water temperature of 20 degrees centigrade, and you remove 500l at a time and heat the water to 90c then return it to the pool, how many litres do you need to heat and return until the pools temperature is at 30c? And how many for 35c?

Thanks for any help

Let us assume that the energy required to heat-up 1000 litre (=1 kL)of fluid by 1 C is H (calorie) and the energy released during cooling down 1kL fluid by 1 C is H.

Let us assume that the Volume of fluid removed (in total) is V (kL). If we assume that the

energy released by hot fluid (90 → 30) with a Volume of V kL = energy absorbed by the cooler fluid (20 → 30) with a volume (20 - V) kL

You can calculate V (kL) from the relationship stated above.

Think about the assumptions we need to make for the process to complete without complications.

Assume:

There is no loss of volume due to evaporation.

The properties of water (density, heat-capacity, etc.) remain constant in the temperature domain (20 - 90 C)

Then (continuing from response #10) :

(20 - V) * H * (30-20) = V * H * (90-30)

200 - 10*V = 60*V

V = 20/7 = 2.8571 kL

Now calculate the same using the equation proposed in #11.