How many kilograms of water can be evaporated at 100 degrees Celsius by the combustion of 470 litres of fuel oil whose heat of combustion is 37 MJ/Litre
To calculate the amount of water that can be evaporated by the combustion of fuel oil, we need to consider the energy released during combustion and the energy required to evaporate water.
First, let's convert the volume of fuel oil from liters to kilograms. We can use the density of fuel oil to do this. The density of fuel oil can vary, but for the purpose of this explanation, let's assume it has a density of 0.95 kg/L. Therefore, the mass of 470 liters of fuel oil can be calculated as:
Mass of fuel oil = Volume × Density
Mass of fuel oil = 470 L × 0.95 kg/L
Mass of fuel oil = 446.5 kg
Next, we need to calculate the energy released during the combustion of fuel oil. Given that the heat of combustion is 37 MJ/Liter, the total energy released can be calculated as:
Energy released = Heat of combustion × Volume of fuel oil
Energy released = 37 MJ/L × 470 L
Energy released = 17,290 MJ
Now, we need to determine the energy required to evaporate water. The heat required to evaporate water is known as the latent heat of vaporization, which is approximately 2,260 kJ/kg.
Finally, we can calculate the amount of water that can be evaporated by dividing the total energy released during combustion by the energy required to evaporate water:
Amount of water evaporated = Energy released / Latent heat of vaporization
Amount of water evaporated = 17,290 MJ / 2,260 kJ/kg
Amount of water evaporated = 17,290,000 kJ / 2,260 kJ/kg
Amount of water evaporated ≈ 7,660 kg
Therefore, approximately 7,660 kilograms of water can be evaporated at 100 degrees Celsius by the combustion of 470 liters of fuel oil with a heat of combustion of 37 MJ/Liter.