Unifying Heat of reaction

Question

Calculate the amount of heat needed to boil 74.1 g

 of water beginning from a temperature of 25.0°C

.

Answer 1

First, find boiling energy (to 100 degrees Celsius)

q=m*c*delta T c=4.184 J/g*C (water's specific heat)

q= 74.1 g x 4.184 x (100-25) = 23,252 J

Now, the heat to get the water from liquid to steam must be calculated with water's enthalpy of vaporization: delta H vap=40.68 kJ/mol

We need to use moles of water

74.1 g /18 g *1 mol = 4.12 moles

q=n x delta Hvap

1= 4.12 moles x 40.68 = 167.6 kJ

167.6 kJ + 23.252 kJ = 190.9 kJ

Answer should have 3 significant figures: 191 kJ