Question

Assuming you have 105grams of water at 23oC, calculate the change in enthalpy required to boil all the water assuming that the enthalpy of vaporization is 40.65kJ/mol and the specific heat capacity is the same as problem

Answer 1

ΔH = 270817.1 J = 270.8 kJ

Step-by-step explanation:

Step 1: Data given

Mass of water = 105 grams

Temperature = 23.0 °C

The enthalpy of vaporization is 40.65kJ/mol

Step 2: Calculate moles H2O

Moles H2O = mass H2O / molar mass H2O

Moles H2O = 105 grams / 18.02 g/mol

Moles H2O = 5.83 moles

Step 3: Calculate the change in enthalpy

ΔH = m*c*ΔT + n(H2O)*Δvap

⇒with ΔH = the change in enthalpy = TO BE DETERMINED

⇒with m = the mass of the water = 105 grams

⇒with c = the specific heat of water = 4.184 J/g°C

⇒with ΔT = the change of temperature = T2 - T1 = 100 °C - 23.0 °C = 77.0 °C

⇒with Δvap = 40.65 kJ/mol = 40650 J/mol

⇒n(H2O) = the moles of H2O = 5.83 moles

ΔH = 105 g* 4.184 J/g°C * 77.0 °C + 40650 J/mol * 5.83 mol

ΔH = 33827.6 J + 236989.5 J

ΔH = 270817.1 J = 270.8 kJ

Answer 2

Enthalpy = 271143.81 J

Step-by-step explanation:

Amount of enthalpy required to boil water = H1 + H2

H1 = MCΔT

H1 = 105 * 4.186 * (100-23)

H1 = 33843.81 J

H2 = M * l

L = enthalpy of vapourization = 40.65kJ/mol = 2260 J g-1 (Upon conversion of units)

H2 = 105 * 2260 J g-1

H2 = 237300J

Amount of enthalpy = 33843.81 + 237300

Enthalpy = 271143.81 J