Final answer:
To boil 100 grams of water at 100°C and 1 atmosphere, a total of 4070 kJ is required, calculated by multiplying the energy needed for 1 gram, which is 40.7 kJ, by 100.
Step-by-step explanation:
The task is to calculate the total number of kilojoules required to boil 100 grams of water at 100°C and 1 atmosphere of pressure. Based on the provided information, when 1 gram of water at 100°C and 1 atm is converted to water vapor at 100°C, 40.7 kJ is absorbed.
Therefore, to calculate the energy needed for 100 grams, we simply multiply the quantity of kilojoules for 1 gram by 100.
Calculation step:
- Energy required to vaporize 1 gram of water = 40.7 kJ
- Total energy required for 100 grams = 40.7 kJ/gram × 100 grams = 4070 kJ
Thus, the total energy required to boil 100 grams of water under the specified conditions is 4070 kJ.