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19.7 kJ
46389000 j
heat of fusion
1 calorie is needed to raise 1 g of water 1 °C. 350 * 22 = 7700 calories ■
41.72 mL, 1.411 oz, or a little under a quarter cup.
If you warm it from 35 degrees Celsius to 1000 degrees Celsius, a mas will vastly increase in volume or pressure. Without knowing how you intend to allow for that, your question is unanswerable.
7.014*10^4
314j
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19.7 kJ
It depends on which temperature you want to reach and at which temperature is the water before you start heating it!
46389000 j
You need to add all of the following:* The heat required to heat ice from -5 to 0 degrees. Multiply the mass times the temperature difference times the specific heat of ice. * The heat required to melt ice. Multiply the mass by the heat of fusion. * The heat required to raiste the temperature of water from 0 to 20 degrees. Multiply the mass times the temperature difference times the specific heat of water.
Assuming standard atmospheric pressure, 2260 kilojoules.
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
It depends on the density of the gel.
heat of fusion