calculate the heat required to change 41.0 g of liquid water at 14.5 C to steam at 100 C and 1 atm. The specific heat of water is 4.184 J/ (g C) and waters enthalpy of vaporization is 2260 J/mol.

Answer 1

q1 = heat needed to raise temperature of liquid H2O from 14.5 C to 100 C,
q1 = mass H2O x specific heat H2O x (Tfinal-Tinitial)
q1 = 41.0 g x 4.184 J/g*C x (100-14.5) = ? J
q2 = heat need to change water @ 100 C to steam @ 100 C
q2 = mass H2O x heat vaporization
q2 = 41.0 g x 1 mol/18.0 g x 2260 J/mol = ?
Qtotal = q1 + q2 = ?
Post your work if you get stuck.
Note: for q2 that 1 mol/18.0 g is to convert the 41.0 g H2O to mols. That is necessary since the heat of vaporization is given in J/mol. It was NOT necessary in q1 since the mass is given in g and the specific heat is given in J/g*C.