How much energy is required to change one quarter of 8.0 kg of water at 25 degree Celsius into steam?
- HomerLv 58 years agoFavorite Answer
takes 1 cal/gm per degree C to get the water to 100 degrees C.
So 1/4 of the 8000 gm is 2000 gm. Then you take the 2000 gm x 1 cal / gm degree x 75 degrees yields units of calories.
Take that value and add to that the result of the following where you have to factor in the heat of vaporization into the equation. The heat of vaporization is the heat needed to vaporize, in this case water, from 100 degrees into steam.
Heat of vapor is 540 cal/gm. Therefore 2000 gm x 540 cal/gm yields units of calories.
Take the initial calorie count from raising the water 75 degrees, and add the calories from the heat of vaporization step gives you a result.
- 4 years ago
i anticipate they choose the respond in energy and you advise the ice is at -12 stages C and atmospheric tension a million calorie = warmth required to develop a million gram of water a million degree celcius the only element i don't have accessible is the table for ice to water and water to steam energies look those numbers up and upload to this equation. 40 5(12 + warmth for melting + one hundred + warmth of vaporization +12) = 40 5*(124 + warmth of melting + warmth of vaporization) =