To calculate the energy required to boil 1 kg of water at room temperature, we need to consider the specific heat capacity and the heat of vaporization of water.
First, we need to heat the water from room temperature to its boiling point. The specific heat capacity of water is approximately 4.18 joules/gram°C (or 4186 joules/kilogram°C). Room temperature is typically around 20°C, and the boiling point of water is 100°C. Therefore, we need to calculate the energy required to heat the water from 20°C to 100°C.
Energy for heating = mass * specific heat capacity * temperature change
Mass of water = 1 kg Specific heat capacity of water = 4186 joules/kilogram°C Temperature change = 100°C - 20°C = 80°C
Energy for heating = 1 kg * 4186 joules/kilogram°C * 80°C Energy for heating = 334,880 joules
Next, we need to consider the energy required to vaporize the water. The heat of vaporization of water is approximately 2,260,000 joules/kilogram.
Energy for vaporization = mass * heat of vaporization
Mass of water = 1 kg Heat of vaporization of water = 2,260,000 joules/kilogram
Energy for vaporization = 1 kg * 2,260,000 joules/kilogram Energy for vaporization = 2,260,000 joules
Finally, we sum up the energy required for heating and vaporization:
Total energy = Energy for heating + Energy for vaporization Total energy = 334,880 joules + 2,260,000 joules Total energy = 2,594,880 joules
Therefore, it takes approximately 2,594,880 joules to boil 1 kg of water at room temperature.