# A 2.15 kg block of ice at 0°C is added to a picnic cooler. How much heat will the ice remove as it melts to wa?

A 2.15 kg block of ice at 0°C is added to a picnic cooler. How much heat will the ice remove as it melts to water at 0°C?

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Phase changes (from solid to liquid for example) happen by the transfer of thermal energy (heat).

For water there is a specific value found in a textbook or table of some sort that tells you how many joules of thermal energy are absorbed as 1 mole of water changes phases from solid to liquid. Take that number (for water its 6.01 kJ/mol) and multiply it by the number of moles of water you have.

In this case, you have 2.15 kilograms which equals 2150 grams.

The molar mass of water is 18 g/mol (this information comes straight from the periodic table of elements).

So: (2150 g) / (18 g/mol) = 119.44 mole of water.

119.44 moles of water x 6.01 kJ/mol = 718.83 kJ of heat absorbed by the ice block to melt.

Source(s): I am a middle school science teacher with degrees in chemistry and physics.
• the warmth of fusion for water is seventy 9.seventy two cal/g. you have 2.5 Kg = 2500g, so which you will pass 2500g * seventy 9.seventy two cal/g = 199,3 hundred energy from the foodstuff to the ice-water by capacity of melting the ice. in case you decide on this in joules, multiply by capacity of four.184 to get 833,900 joules. those kinds of numbers are rounded to 4 places, yet you could around in any different case in the adventure that your accuracy replaced into specific in any different case.

• Heat of fusion= mass X heat of fusion, 80cal/g heat of fusion for ice

Heat= 2150 g X 80 cal/g= 172,000cal or 172Kcal