Question

Calculate the energy (in kJ) required to heat 11.8 g of liquid water from 55 oC to 100 oC and change it to steam at 100 oC. The specific heat capacity of liquid water is 4.18 J/goC, and the molar heat of vaporization of water is 40.6 kJ/mol.

Answer:

Answer 1

Step-by-step explanation:

First, heat it from 55-100 C

11.8 g ( 100 - 55) C 4.18 J / g C = 2219.6 J = 2.22 kJ

Your heat of vaporization is in units of moles

so 11.8 g of H2O = 11.8 gm / 18 gm /mole = .656 moles

Then

.656 moles * 40.6 kJ / mole = 26.6 kJ

Total kJ = 2.22 + 26.6 = 28.8 kJ