the specific heat of ice Cm(ice.=2.110 J/(g*C) The specific heat of liquid water
ID: 825110 • Letter: T
Question
the specific heat of ice Cm(ice.=2.110 J/(g*C)
The specific heat of liquid water,Cm liquid water,Cm (liquid)=4.184J(g*C)
The specific heat of water vapor,Cm vapor=2.080 J/(g*C)
Heat of fusion for water H2O 6.01 kj mol
Heat of evaporation for water deltaH evap=40.8 kj/mol
Density of liquid water assume 1.00 g/ml
Molecular weight 18.02 g/mol
How much heat is required in is required to bring 250 ml of water from 25 degrees to its boiling point temperature 100 C.?
How much is required evapoate 250 ml of water at its boiling point temp 100 celcius (at 1 atm)?
Explanation / Answer
(1) heat required to bring 250 ml of water from 25 degrees to its boiling point temperature 100 C = mass of water * The specific heat of liquid water * change in temperature
now, density of water = 1g/ml
thus,mass of water =250 g and change in temperature = (100 -25)=75
therefore, heat required to bring 250 ml of water from 25 degrees to its boiling point temperature 100 C =250*4.184*75 = 78450 J = 78.45 kJ
(2) heat required to evapoate 250 ml of water at its boiling point temp 100 celcius (at 1 atm) = mass of water * Heat of evaporation for water
now, mass of water =250 g = 250/18 moles = 13.89 moles
Heat of evaporation for water = 40.8 kJ/mol
therefore,heat required to evapoate 250 ml of water at its boiling point temp 100 celcius (at 1 atm) =13.89*40.8 =566.67kJ
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.