How much energy is required to turn 150 grams of water into steam at a temperature of 100 degrees?

The required amount of energy for water vaporization is calculated by the formula:

Q = L * m, where L is the specific heat of vaporization of water (L = 2.3 * 10 ^ 6 J / kg), m is the mass of water (m = 150 g = 0.15 kg).

Let’s calculate the required amount of energy:

Q = L * m = 2.3 * 10 ^ 6 * 0.15 = 345000 J = 345 kJ.

Answer: To vaporize 150 g of water, you need to spend 345 kJ of thermal energy.