1. The heat of vaporization of water is 40 kJ/mol at 100°C. What percentage of the heat supplied to vaporize the water can be converted into work if the water is vaporized at 100°C at a constant pressure of 1.0 atm?

2. A block of lead (specific heat=0.128 J/K*g) at 90°C is dropped into water at 20°C in a well insulated calorimeter whose heat capacity is 131 J/K. The final temperature of the water and the lead block is 22°C. What is the mass of the lead block?

Heat required to convert 1 mole of water to steam =40KJ/mole
Work is done by the system because when water is converted into heat ,expansion takes place.
w= P(V2-V1)
density of water =1g/ml
molar mass of water = 18g/mole
V1 is the volume of 1mole water at 1atm = mass/density= 18g/1g/ml =18ml = 18 X 10^-3 dm^3
V2 is the volume of 1mole of water at 100 deg C
V 2 = 22.4 dm^3 X 373K /273K  = 30.60 dm^3

P = 1 atm = 101,325N/m^2(pa)
w = 101,325(30.60- 0.0018) X 1m^3/ 1000dm^3= 3100J = 3.1KJ
now percentage = 3.1KJX 100%/ 40KJ = 7.75%

2. A block of lead (specific heat=0.128 J/K*g) at 90°C is dropped into water at 20°C in a well insulated calorimeter whose heat capacity is 131 J/K. The final temperature of the water and the lead block is 22°C. What is the mass of the lead block?


Heat absorbed in calorimeter = heat capacity X rise in temperature of water  = 131 X2 =262 J

Heat absorbed = mC delta t
or m = heat absorbed  /(C X change  in temp of lead)  = 262 / (0.128 x 70 ) = 29.24 g