How much thermal energy is necessary to vaporize 36.000 g of water at its boiling point?

To calculate the thermal energy required to vaporize water, we use the formula: ( Q = m \times L_v ), where ( m ) is the mass of the water and ( L_v ) is the latent heat of vaporization of water, approximately 2260 J/g. For 36,000 g of water, the thermal energy needed would be ( Q = 36,000 , \text{g} \times 2260 , \text{J/g} ), which equals 81,360,000 J, or 81.36 MJ. Thus, 81.36 megajoules of thermal energy is necessary to vaporize 36,000 g of water at its boiling point.