1. **State the problem:** We need to find the energy required to vaporize 81.37 g of water at 100°C. 2. **Formula used:** The energy required for vaporization is given by $$Q = m \times L_v$$ where $Q$ is the heat energy, $m$ is the mass of the substance, and $L_v$ is the latent heat of vaporization. 3. **Known values:** - Mass of water, $m = 81.37$ g - Latent heat of vaporization of water at 100°C, $L_v = 2260$ J/g (joules per gram) 4. **Calculate the energy:** $$Q = 81.37 \times 2260 = 183,996.2 \text{ J}$$ 5. **Interpretation:** The energy needed to vaporize 81.37 g of water at 100°C is approximately 184,000 joules. This means you must supply about 184,000 joules of heat to convert 81.37 g of water at boiling point into steam without changing the temperature.