What amount of heat, in kJ, is required to convert 1.00 g of water at 67.0 °C to 1.00 g of steam at 100.0 °C? (specific heat capacity of water = 4.184 J/g • °C; ∆Hvap = 40.7 kJ/mol)

Question

What amount of heat, in kJ, is required to convert 1.00 g of water at 67.0 °C to 1.00 g of steam at 100.0 °C? (specific heat capacity of water = 4.184 J/g • °C; ∆Hvap = 40.7 kJ/mol)

Answer 1

two steps are required here. first step is to figure out how much energy is required to heat the water from 67 to 100. the second step is to figure out how much energy is required to convert all the water (at 100) from a liquid to a gas

step 1: Q=MCΔT

Q= (1g)(4.184j/g)(100-67)

138.072 J = 0.138072 kJ

step 2: Q=MΔHvap

ΔHvap was given in units of moles so we have to convert 1g water into moles of water

1g*(1mol/18.02g)= .05549 mols

Q= (.05549mol)(40.7kj/mol)

2.2586 kJ

last step is to add the heat from steps 1 and 2

0.138072 +2.2586= 2.397 kJ