A piece of copper of mass 300 g at a temperature of 950 degree Celsius is quickly transferred to a vessel of negligible thermal containing 250 g of water at 25 degree Celsius. If the final temperature of the mixture is 100 degree Celsius calculate the mass of the water that will boil away ( specific heat capacity of copper =400 j/kg.k specific latent heat of vaporization of steam =2260000jkg

9 answers

0.300 * 400 * (950-100) = 102,000 Joules lost by copper

that heats 0.250 kg water to 100 from 25
0.250 * 4186 * 75 = 78488 Joules used from copper to heat water.
That leaves 23,512 Joules to boil water
23,512 = m 2,260,000
m = .0104 kg = 10.4 grams
Thanks
What if the special heat capacity of water is 4.2×10^2
Where is the 4186 from
Yes
Where is the 4186 from? i am confused
That not the answer
i dont even understand this solving sef
I don't understand the solving