The answer will depend, among other things, on the calorific value of the fuel.
The answer is 900 kg.
1 kg
latent heat
4 kg is the mass of the water. Thus, 4 kg of water has a mass of 4 kg.
173 kg pure water ie equivalent to 173 L.
To evaporate water you must supply the latent heat of vaporisation. This is large for water compared with many other liquids. It is 2,260,000 joules per kg.
Anything will float on water as long as it's density is lower than that of water. The density of jet fuel is around 0.81 kg/L while that of pure water is 1 kg/L. Thus, jet fuel floats.
do the math
1,400 kg of water is 6,172.6 cups1,400 kg of water is 6,172.6 cups
3.785 kg
187 US gallons is approximately 578 Kg.
It entirely depends on how much power or rate of energy flow there is into the water, which requires the latent heat of vaporisation (2.26 x 106 Joules/kg for water) to be supplied. Thus if the energy flow is 1 kW = 1000 Joules/sec, the mass of water evaporated per second will be 1000/2.26 x 106 kg or 0.4425 x 10-3 kg, which means it will take 2260 seconds to evaporate one kg, or about 38 minutes. Obviously in a large power plant where the energy flow might be 1000 MW, this is one million times as much, and you would evaporate 1000 tonnes (a million kg) in the same time.
135 kg of water
The answer is 900 kg.
73440
Fossil fuels are hydrocarbons, so the main processes are: 1. carbon (fuel) + oxygen (air) produced carbon dioxide and heat 2. hydrogen (fuel) + oxygen (air) produces water (usually steam) and heat. In a typical fuel-burning process, 1 kg of fuel burns with 3.5 kg of oxygen to produce 3.1 kg of carbon dioxide and 1.4 kg of water. This is inescapable because the heat energy is released by the formation of the carbon-oxygen and hydrogen-oxygen chemical bonds.
1 ton = 1000 kilograms so 160 tons = 160,000 kg. Simple!