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Since you want your answer in grams of ice we need to convert 351 ml of water to grams. This is very easy for water since 1 ml of water weighs one gram. So we have 351 grams of water.

We need to figure out how much energy we need to remove from the water. If you remove one calorie from one gram of water you will lower the temperature one degree C. So water has a specific heat (c) of 1 cal/g/C

You have 351 grams and you want to drop the temp 20 degrees. You must remove

q = mass x specific heat x temp change

q = 351 grams x 1 cal/g/C x 20 degrees= 7020 calories

now we need to figure out how much ice this will take.

Melting one gram of ice requires 79.7 calories dividing we get

ice required = 7020 calories/79.7 calories/grams = 88.1 grams of ice

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14y ago
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14y ago

I had to get my old lab book out to answer this. An enthalpy of fusion problem.

You have 1 gram/ml density of water ( I know, ice is less dense, but I am too lazy to look it up ) so you have 355 grams of water. Since you are in grams and want grams; 334 Joules per gram is Heat of fusion for water. Use q absorbed + q evolved = 0 for this problem. Set up.

(grams)(334J/g) + (grams)(4.18J/gC)(6C-0C) + (355g)(4.18J/gC)(6C-25C) = 0

= 78.5 grams of water

( I know. I am too lazy to do the math. Math provided by TI-84 equation solver function. )

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14y ago

*Any* amount will lower it.

You need to specify the temperature drop needed.

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Q: How much ice would have to melt to lower the temperature of 352 ml of water?
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