# How many grams of ice at -14 degrees Celsius are needed to cool 200 grams of water from 25 degrees Celsius to 10 degrees Celsius?

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The heat lost by the water in cooling to 14 degrees C is equal to the heat gained by the ice.

The heat capacity "c" of water, 4.18 J/g-degree C, is the amount of energy needed to heat one gram of water one Celsius degree. It's also the amount of heat released when one gram of water is cooled one Celsius degree. To find the amount of heat lost by 200 grams cooling from 25 to 10 degrees C we use:

q = mc `Delta` T = (200g)(4.18 J/g-deg C)(10 C) = 836 Joules

There are three steps to the ice absorbing heat:

1. it warms from -14 to 0 deg C

2. it melts

3. it warms from 0 to 10 deg C

The heat capacity "c" of ice is 2.03 J/g-deg C

The heat of fusion (heat required to melt ice) is 334 J/g

With x representing the grams of ice, the equations for the three steps are:

1. q = mc `Delta` T = (x)(2.03 J/g-deg C)(14 deg C) = 28.42x J

2. q = (m)(heat of fusion) = (x)(334J/g) = 334x J

3. q = mc `Delta` T = (x)(4.18 J/g-deg C)(10 deg C) = 41.8x J

The total heat gained by the ice, in terms of x, is:

28.42x + 334x + 41.8x = 404.22x

The heat lost by the water initially at 25 degrees equals the heat gained by the ice, so:

404.22x J = 836 J

x = 2.07 grams of ice