When water is confined under pressure greater than normal atmospheric pressure,
a higher temperature is required to make it boil. Higher pressure can be created
simply by sealing the container the water is in ... that's how pressure cookers work.
Just seal the water in a container that can stand the pressure required to raise the
boiling point to 1,000 degrees, keep adding heat, and you'll get there.
NOTE:
DO NOT try this at home. You don't know what pressure is required, you
don't know how much pressure your container can stand, and you have no
way to measure the temperature of the water in it. If you put water in a
sealed container and heat it, the only thing that can be guaranteed for sure
is that your container will explode, and the stronger the container is, the more
violent the explosion will be.
To heat 1000 grams of water from 20°C to 80°C, you need to raise the temperature by 60°C. The specific heat capacity of water is approximately 4.18 J/g°C, so the energy required is 1000 g × 4.18 J/g°C × 60°C = 250,800 J. A 1 kW kettle provides 1000 J/s, so it will take about 250.8 seconds, or approximately 4.2 minutes, to heat the water.
To raise 1000 grams of water from 50 to 100 degrees requires 50 degrees x 1000 grams of heat, so the answer is 50,000 calories. Water at 100 degrees requires an additional 550 calories to convert 1 gram fully into steam. Therefore the remaining 50,000 calories can convert 50,000/550 grams into steam. So 90.9 grams become steam, and that's the answer.
No, it measures the ammount of heat output that would heat a millilitre of water 1 degree Celsius. So if you were to burn 1000 calories of energy, it would heat one millilitre 1000 degrees Celsius or it would heat 1 litre 1 degree Celsius.
100 degrees celsius are equal to 212 degrees fahrenheit.
YES it will demagnitize at round 800 to 1000 degrees
Water boils at 100C (or 212F) at sea level.
a lot
The heat required to evaporate 1 liter of water at 100 degrees Celsius is known as the latent heat of vaporization of water, which is approximately 2260 kJ/kg. Since the density of water is about 1000 kg/m³, the heat required would be around 2260 kJ.
Heat required for this transition is given as the the sum of three heatsheat required for heating the ice from -5 degree Celsius +latent heat(conversion of ice at zero degree to water at zero degrees)+heat required to heat the water from 0 to 5 degree CelsiusHeating of ice=m x s x delta T,where m is the mass ,s is the specific heat of ice=200x0.5x5=500calmelting of ice=mxlatent heat=200x80=16,000calHeating of water=m x s x delta T,where m is the mass ,s is the specific heat of water =200x1x5=1000calTotal heat required=500+16,000+1000=17,500 cal
Water boils at 100 degrees Celsius or 212 degrees Fahrenheit.
The specific heat capacity of water is 4.186 J/g°C. Since there are 1000 grams in a kilogram, it would require 20,930 Joules of energy to increase the temperature of a kilogram of water by 5 degrees Celsius.
To heat 1 gram of water by 1 degree Celsius, it takes 4.18 joules. So, to heat water from, for example, 20 degrees to 100 degrees, you would need to calculate the total mass of water and apply the specific heat capacity to determine the total energy required.