How long would it take for 1.50 mol of water at 100.0 ∘C to be converted completely into steam if heat were ad?
How long would it take for 1.50 mol of water at 100.0 ∘C to be converted completely into steam if heat were added at a constant rate of 20.0J/s ?
given:
The constants for H2O are shown here:
Specific heat of ice: sice=2.09 J/(g⋅∘C)
Specific heat of liquid water: swater=4.18 J/(g⋅∘C)
Enthalpy of fusion: ΔHfus=334 J/g
Enthalpy of vaporization: ΔHvap=2250 J/g
More Questions From This User See All
Answers & Comments
Verified answer
The mass of the water m is 1.50 mol x 18g/mol=27g
As in the last problem you need to use the formula Q=m x Enthalpy of vaporization to calculate the energy needeed to break up the ties among atoms: Q1=27g x 2250 J/g
Now you can calculate the time t with this formula: t=Q1 / 20J/s
Ehi i'm not even english and the best answer was mine last time
I know this post has been posted for quiet long but I tried the answer that lukel provided and it was not right. So I worked it out.
We have H(vap) = 40.67J/g*C
so q=(40.67 kJ/g*C)(1.50 mol) = 61.005 kJ (3sF) = 61.0 kJ (recorded value)
Heat was applied at a constant 20.0J/s = 0.0200 kJ/s
Therefore, 61.005 kJ/0.0200 kJ/s = 3050.25 s
= 50.8375 min (record 50.8 min)
or = 0.847292 hr (record 0.847 hr)
50.6 min