For the heating curve of water at 1.00 atm, is the slope associated with heating
ID: 492393 • Letter: F
Question
For the heating curve of water at 1.00 atm, is the slope associated with heating liquid water greater than or less than that associated with heating ice? Explain your answer by explicitly discussing the relative heat capacities of ice and liquid water. Using the fact that the heat capacity of ice is 2.09 J.g^-1.0 C^-1, the heat capacity of liquid water is 4.184 J middot g^-1.0 C^-1, and (for water) delta H_dis = 6.02 kJ/mol, and delta H_tap = 40.7 kJ/mol, determine the heat required to convert 50.0 mL of water at 22.5 degree C (assume a density of water of 1.0 g/mL) to steam at 100.0 degree C.Explanation / Answer
(3) Q = heat change for conversion of water at 22.5oC to water at 100 oC +heat change for conversion of water at 100 oC to vapour at 100 oC
Amount of heat absorbed , Q = mcdt + mL
= m(cdt + L )
Where
m = mass of water = volume x density = 50.0 mL x 1.0 g/mL = 50.0 g
c = Specific heat of water = 4.186 J/g degree C
L = Heat of Vaporization of water = 40.7 (kJ/mol )x(1000 J/kJ) x ( 1 mol/ 18 g) = 2261.1 J/g
dt = 100 - 22.5 =77.5 oC
Plug the values we get Q = m(cdt + L ) = 129.3x103 J = 129.3 kJ