They’re equally as hot at 50˚C each. However the one with the larger water mass has the most heat energy, as every gram of water at the same temperature contains the same amount of heat energy.
To have zero heat energy a substance has to be at absolute zero. This is 0˚K which equals -273˚C. So there’s already a lot of heat energy in a substance at 0˚C. The amount of heat energy per degree per gram varies depending on the temperature, but in rough terms a gram of ice contains 2.11 Joules for every ˚C above -273˚C. So a gram of ice at 0˚C contains roughly 273*2.11 = 576 Joules of heat energy.
It then takes an additional 334 Joules to melt a gram of 0˚C ice into a gram of 0˚C water. After this a gram of water requires 4.19 Joules for every ˚C above 0˚C, and so heating this water up to 100˚C then requires 100\times 4.19=419 Joules per gram. So the total heat energy in a gram of 100˚C water is 576+334+419=1329 Joules.
Putting this together means that a gram of boiling water contains 1329/576 = 2.3 times as much heat energy as a gram of 0˚C ice, or put another way, 2.3 grams of ice contains the same heat energy as 1 gram of boiling water.
So in answer to the question if you freeze 2.3 cups of water to just 0˚C, the resulting ice will contain roughly the same heat energy as 1 cup of water at 100˚C. Ice out at sea may of course be a few degrees cooler and so you’d need a little more, but you certainly don’t need anything like an iceberg!
5.7K viewsView upvotes
2
Related Questions (More Answers Below)
