Answer:
The heat required to evaporate 512 g of water initially at 0°C is 1,371,340.8 J
Step-by-step explanation:
The given parameters are;
The mass of water given in the question = 512.0 g
The specific heat capacity of water = 4.184 J/(g·°C)
The heat of vaporization of water = 2,260 J/g
The heat, Q, required to raise the temperature of 512.0 g from 0°C to 100°C is given as follows;
Q = 512 × 100 × 4.184 = 214,220.8 J
The heat required to vaporize 512.0 g of water at 100°C = Mass × Heat of Vaporization
The heat required to vaporize 512.0 g of water at 100°C = 512 × 2260 = 1,157,120 Joules = 1,157.12 kJ
The heat required to evaporate 512 g of water initially at 0°C is therefore;
214,220.8 J + 1,157,120 J = 1,371,340.8 J.