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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 Ive 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 Im pretty stuck since I cant equate Q = mLf and Q = mcΔt to mLf = mcΔt since that would mean the mass is 0, which is incorrect. Im 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.

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