Calculate the Mass of Ice cubes needed to cool a soft drink

• CyanPowder
In summary: Soapstone CupThe mass of Soapstone (MS) needs to increase it's temperature from -5C to 7CWater Soda needs to decrease it's temperature from 17C to 7CThe specific heat of the mass of soda is the same as the specific heat of water (SHC) so the MS will increase its temperature by the same amount as the SHC of water (980 J/kg C).
CyanPowder
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Summary:: TLDR : Drink cooled from 17 C to 7 C with either ice cubes (method 1) or soapstone stones (method 2). Calculate the mass of each item that would be needed to cool the same drink. Given the information provided.

In order to cool a drink (“cola” for example) from a storage-room temperature of 17.0 C to a more pleasant “drinking” temperature of 7.0 C, we often put ice cubes in the drink called method I. A product made of soapstone is sold as an alternative to ice cubes --- “soapstone rocks” of are first chilled in the freezer, and then the rocks are placed in the drink called method II.

Assume:
Freezer temperature is – 5.0 C with lots of ice cubes and “soapstone rocks”
SHC of the soapstone is 980 J/kg C
LFC of water 334 kJ.kg
SHC of water 4184 J⋅kg −1 ⋅K −1
Mass of cola to be cooled is 150 g (about a ½ can of pop)
Soapstone density is ~2.956g/cm3

In an attempt to get the final temperature of the “cola” to 7.0 C both methods are compared. Calculate the mass of each method of cooling required and justify and explain your answer.

So far what I've done is calculate Δt for both the "cola":

Δt = 7.0C - 17.0C = -10.0C

and ice cubes:

Δt = 7.0C - (-5.0C) = 12.0C

and converted LFC of water from kJ/kgK to J/kgK: 334 kJ/kg to 334000 J/Kg.

At this point I am pretty stuck since I can't equate Q = mLf and Q = mcΔt to mLf = mcΔt since that would mean the mass is 0, which is incorrect. I am aware that the ice has to change from solid state to liquid state and that total energy could be calculated by adding (mLf) + (mcΔt), but that still leaves m as an unknown, I have absolutely no clue where to go from here.

Last edited:
Is soapstone ice-cubes a real thing ? curious.

Anyways, forum rules require that you show your work ; what you've done, so far.

hmmm27 said:
Is soapstone ice-cubes a real thing ? curious.

Anyways, forum rules require that you show your work ; what you've done, so far.
It's two different things that can be used interchangeably to cool a drink, soapstones are one, and ice cubes are another, and thank you for the telling me about the "show what you've done so far", I just joined today so I had no idea.

CyanPowder said:
Summary:: TLDR : Drink cooled from 17 C to 7 C with either ice cubes (method 1) or soapstone stones (method 2). Calculate the mass of each item that would be needed to cool the same drink. Given the information provided .

At this point I am pretty stuck since I can't equate Q = mLf and Q = mcΔt to mLf = mcΔt since that would mean the mass is 0,
I would say you are pretty well there. What you need is the good old "method of mixtures" which we did in the 50's. The 'total Energy' would be the same on both sides of a before and after equation. Putting the total masses X SHC and the mass of ice X Latent Heat on the appropriate sides will involve just one unknown quantity ΔT.
For the cold stones, you just have masses X SHC on both sides to consider.
The Energy can be chosen to be relative to any temperature - say 0C.
I'm sure you can do it which what you've already shown us. (If you are still flummoxed, try writing a 'word equation' first.)

Lnewqban
CyanPowder said:
It's two different things that can be used interchangeably to cool a drink
They are not interchangeable from a functional point of view. One dilutes the beverage and the other doesn’t.

This is how I would do it...

We know that the final temperature of the objects in the cups must be the same (7C).
A "conservation of energy" type approach is usually the way these types of problems are done from memory.

Note: Heat Energy (or change in Heat Energy)= Q

There is a fair bit of information in this problem that needs to be managed and so structure is important- this is what trips a lot of people up. The calculations are trivial.

The questions you can ask are- what do I know about the systems- what assumptions do I need to make- what equations do I need to use - there may be red herrings in the question that can trip you up (ie. you don't really need to know the Density of Soapstone)- if you are methodical it will help- and if the first method doesn't work you can start again. All comes down to good maths hygienne.Soapstone Cup

The Mass of Soapstone (MS) needs to increase it's temperature from -5C to 7C
Water Soda needs to decrease it's temperature from 17C to 7C

Assume the specific heat of the mass of soda is the same as the specific heat of water.(Change In Heat) Q Water Soda = Q SoapstoneIce Cup

Ice needs to increase it's temperature from -5C to 7C including the Latent Heat Effects of the transition from Ice to Water.

Assume the specific heat of the mass of ice (MI) is the same as the specific heat of water.

(Change In Heat) Q Ice= Specific Heat Change + Latent Heat Change

Q Water Soda (already calculated above) = Q Ice

Check that the answers are reasonable- Ice has four times the Specific Energy of Soapstone so this should be reflected in the MI:MS ratio. The effect due to Latent Heat will be much smaller than due to Specific Heat (because of the coefficients) and so won't have as much influence over the result.

Last edited:
Orodruin said:
One dilutes the beverage and the other doesn’t.
The obsession with chilled drinks, these days, (chocolate too!) has led to spoiling a lot of tastes.

And if you want a cold drink then keep all your bottles in a fridge. Vodka can be drunk quickly (in small doses) and a frequent top-up from the fridge gives a better result than a tepid, half empty glass as you wade through it.

In the case of cheapo whiskies, you need at least some water to take the edge off them so melted ice does the job. Personally I'd rather have fewer drinks of good single malt than a load of nasty blended stuff.

But that's not Physics so it's off topic.

CyanPowder said:
Assume:
Freezer temperature is – 5.0 C with lots of ice cubes and “soapstone rocks”
SHC of the soapstone is 980 J/kg C
LFC of water 334 kJ.kg
SHC of water 4184 J⋅kg −1 ⋅K −1
Mass of cola to be cooled is 150 g (about a ½ can of pop)
Soapstone density is ~2.956g/cm3

In an attempt to get the final temperature of the “cola” to 7.0 C both methods are compared. Calculate the mass of each method of cooling required and justify and explain your answer.

So far what I've done is calculate Δt for both the "cola":

Δt = 7.0C - 17.0C = -10.0C

and ice cubes:

Δt = 7.0C - (-5.0C) = 12.0C

and converted LFC of water from kJ/kgK to J/kgK: 334 kJ/kg to 334000 J/Kg.

At this point I am pretty stuck since I can't equate Q = mLf and Q = mcΔt to mLf = mcΔt since that would mean the mass is 0, which is incorrect. I am aware that the ice has to change from solid state to liquid state and that total energy could be calculated by adding (mLf) + (mcΔt), but that still leaves m as an unknown, I have absolutely no clue where to go from here.
Why don't you start with the soapstone then, since there's only hc to deal with.

(and, apologies for taking so long to get back to you : apparently the forum decided not to notify me of this thread updates until this morning)

Orodruin said:
One dilutes the beverage
Only if you don't put enough in (or drink it too slowly). This is often addressed in a follow-up question for extra marks.

Edit: I just reread the original question: if you are only cooling to 7°C then you are not putting enough in.

“Shaken, not stirred” with small pieces of ice, then strained when the shaker feels cold enough. No surplus water this way - just the needed amount of melt. The fashion is to have cans in a cool box of icy water (or even salt freezing mixture.
You can get cubes with freezing mixture inside, too. No melt water at all

I like the idea of soapstone cold-cubes, but wouldn't it tend to break glass cups ?

1. How do I calculate the mass of ice cubes needed to cool a soft drink?

To calculate the mass of ice cubes needed to cool a soft drink, you will need to know the initial temperature of the drink, the desired temperature, and the specific heat capacity of the drink. You can then use the formula: mass = (specific heat capacity x volume x change in temperature) / specific heat capacity of ice. This will give you the mass of ice cubes needed in grams.

2. What is the specific heat capacity of ice?

The specific heat capacity of ice is 2.09 J/g°C. This means that it takes 2.09 Joules of energy to raise the temperature of 1 gram of ice by 1 degree Celsius.

3. Can I use any type of ice cubes to cool a soft drink?

Yes, you can use any type of ice cubes to cool a soft drink. However, the size and shape of the ice cubes may affect the rate of cooling and therefore the amount of ice cubes needed.

4. How do I know the specific heat capacity of a soft drink?

The specific heat capacity of a soft drink can vary depending on the type and brand. You can usually find this information on the label of the drink or by searching online. Alternatively, you can use an average value of 4.18 J/g°C for most soft drinks.

5. Can I use this formula to calculate the mass of ice cubes for any type of drink?

Yes, you can use this formula to calculate the mass of ice cubes needed for any type of drink as long as you know the specific heat capacity of the drink. However, keep in mind that the specific heat capacity may vary for different types of drinks, so the amount of ice cubes needed may differ.

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