a piece of copper of mass 300g at 950 degrees celcius is quickly transfered to vessel of negligible thermole capacity containing 250g of water at 25 degrees celcious. if the final steady temp of the mixture is 100degrees, calculate the masd of water that will boil away

Answer 1

The copper is cooled from 950 to 250
so heat out of copper = 300 *Ccopper (950 -100)
that heat goes into the water which of course ends up at 100 deg if boiling.
heat into water = 250 Cwater (100 -25) + Mass boiled* heat of vaporization of water
solve for Mass boiled.

Answer 2

what is this,you are very stupid,where is your answer

Answer 3

this is rubbish