The amount of energy to boil water depends on two things: the amount of water and the temperature of the water when you start.
The more water water, the more energy it takes. The colder the water, the more energy it takes.
Let say the water is at room temperature, or 20 °C. First, the water must be heated to 100 °C, which takes energy. The amount of energy is given by the specific heat of water, which is 4.186 Joule/gram °C. That means that requires 4.186 Joules of energy to heat 1 kilogram of water by 1 °C. So if you have 1 kilo grams of water at 20 °C, you have to add this much energy:
= (4.18Joule/gram °C) (100g) (100 °C - 20 °C)
= 334400 joules/°C
Of course, if you had more or less water, or it was colder or warmer, you would adjust this equation accordingly..
Water boils at 100 degrees Celsius at standard atmospheric pressure. To boil water at 95 degrees Celsius, you can increase the pressure in a sealed container, which raises the boiling point. Similarly, to boil water at 105 degrees Celsius, you would need to decrease the pressure in a sealed container to lower the boiling point.
The water boiling point in Celsius is 100 °C.
The boiling point of water is 100 degrees Celsius or 212 degrees Fahrenheit. However, water boils at a lower temperature at higher altitude. Salt water boils at a higher temperature than pure water.
Mercury will boil last, as it has a higher boiling point than water and alcohol. Mercury boils at around 674 degrees Fahrenheit, while water boils at 212 degrees Fahrenheit and alcohol at about 173 degrees Fahrenheit.
The heat required to boil water can be calculated by multiplying the mass of water (21.1 g) by the specific heat capacity of water (4.18 J/g°C) and the temperature change (100°C - initial temperature). This calculation results in 8.82 kJ or 8820 J of energy needed to boil 21.1 g of water at 100°C.
The amount of heat needed to boil water varies with the air pressure. Water in a vacuum will boil at room temperature.
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To calculate the energy needed to boil 90 grams of water, we use the formula ( q = m \cdot L ), where ( q ) is the heat energy, ( m ) is the mass of the water, and ( L ) is the latent heat of vaporization for water, which is approximately 2260 kJ/kg. For 90 grams (0.09 kg), the calculation is ( q = 0.09 , \text{kg} \cdot 2260 , \text{kJ/kg} ), resulting in about 203.4 kJ of energy needed to boil the water.
Water reaches a roaring boil at 212 degrees Fahrenheit or 100 degrees Celsius.
If its in Celsius then another 13 degrees are needed because water boils at 100 degrees Celsius
No, since there is no such thing as degrees "celius". Water will boil at 97.0 degrees Celsius at approx 850 metres above sea level.
If by "boil" you mean have it all evaporate, that takes MUCH more energy. For example, to increase the temperature of one gram of water from 20 to 100 degrees Celsius, you need 4.2 joules/gram/degree times 80 degrees = about 336 joules; then, to evaporate all the water, you need an additional 2257 joules.
100 degrees Celsius 212 degrees Fahrenheit
Water boils at 212 degrees Fahrenheit.
Water boils at 100 degrees Celsius or 212 degrees Fahrenheit
The energy required to boil 100 ml of water at room temperature (20°C) to boiling point (100°C) is about 4200 joules. This is the amount of energy needed to raise the temperature of water by 1°C per gram.
Water can boil below 100 degrees Celsius depending on the area of the world. A lab was conducted in science class that our water boiled at 94.6 degrees Celsius. The average boiling point for water is 100 degrees Celsius but that does not mean it will always be that degree to boil.