1. The problem is to calculate the heat energy of the water. 2. The formula to calculate heat energy is $$Q = mc\Delta T$$ where: - $Q$ is the heat energy (in joules), - $m$ is the mass of the water (in kilograms), - $c$ is the specific heat capacity of water (approximately 4184 J/kg°C), - $\Delta T$ is the change in temperature (in °C). 3. To solve, you need the values of $m$ and $\Delta T$. For example, if $m=2$ kg and temperature changes from 20°C to 80°C, then $\Delta T = 80 - 20 = 60$°C. 4. Substitute the values into the formula: $$Q = 2 \times 4184 \times 60$$ 5. Calculate the multiplication: $$Q = 2 \times 4184 \times 60 = 502080$$ 6. Therefore, the heat energy of the water is $$502080$$ joules. This calculation assumes no heat loss to the environment and pure water.