1.750 kg of ice at 0.00 degrees celcius is added to 500.0 g of water in an insulated container. What is the mass of ice that remains unmelted after equilibrium is established?
2007-01-31 01:31:48 · 1 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
I think you need more information. It seems that the 500.0 g water is at some higher temperature like 20 or 25degC. If it's 25degC, then you need to cool the water down to 0degC. That takes 500g x 25deg x 1cal/deg-g = 12,500 cal. Next, you have to get that number of cal by melting some ice. The heat of fusion of ice is 79.8cal/g. 12,500cal/79.8cal = 156.6g ice. The amount of ice remaining is then 1750g 0 - 156.6g = 1593g = 1.593kg.
You have to plug in your own beginning temoperature for water, but that's how you do it.
2007-01-31 02:27:08 · answer #1 · answered by steve_geo1 7 · 0⤊ 0⤋