Energy Needed For Ice to Be Turned Into Steam (Specific Heat and Latent Heat of Fusion/Vaporization
We have a beaker filled with .069 kilograms of ice We want to find out how much heat energy in joules it will take to turn the ice with intial temperature of -5 degrees Celsius into steam at 110 degrees Celsius For the first step of this problem we need to find out how much heat it takes to raise the .069 kilograms of ice to the freezing point of zero degress Celsius. The equation for finding the heat energy needed to raise the temperature of ice is specific heat times mass of ice times the change in temperature. We can plug in our specific heat or how much energy it takes to raise 1 kilogram of ice 1 degree Celsius into the equation with the mass of ice in kilograms with the change in temperature which is 0 degrees Celsius final minus – 5 degree celsisus intial. We get a heat added of 724.6 joules. Now the ice will have to under go a phase change from ice to liquid water. This will require energy. We will need to know what the latent heat of fusion is which is the amount of energy per kilogram of ice to change from a solid to a liquid. The latent heat of fusion is 3.33 times 10 to the 5th joules per kilogram of ice. After multiplying the latent fusion times the mass of ice we get 22,977 additional joules of heat needed to convert the ice to a liquid. For the next step we need to find out how much heat energy will be required to raise the water from zero degrees to 100 degrees. We will use the specific heat for liquid water and multiply it times the mass of the ice which is now water times the heat delta of 100 degrees. We get an additional 28,883.4 joules of heat energy needed to raise the water up to the boiling point. The water will need to change phases again this time from liquid to a gas. So we will need to know what the latent heat of vaporization is. The latent heat of vaporization of water is 22.6 times 10 to the 5th joules a kilogram. This value means that for every kilogram of water it will require 22.6 times 10 to the 5th joules of energy to convert it to a gas vapor. So we can multiply this times how many kilograms of water we have which is .069 kilograms. We get an additional 155,940 joules of heat needed to get the gas to turn into a vapor. Now that gas will have a slightly higher temperature then the boiling point. In this case it has a temperature of 110 degrees Celsius. So we need to look up the specific heat for water when it is a steam. The specific heat is 2,000 joules per kilogram of steam raised by 1 degree celsisus. So now if we multiply the specific heat times the mass of steam and the change in temperature or delta of 10 degrees Celsius we get 1,380 joules of heat needed to raise the gas vapor by 10 degrees Celsius. To finish of this video lets find the total joules of heat needed to get water from ice to steam. We can add up all of the joules of heat needed for the various temperature changes and phases changes. So after we added up q1 through 5 we get an answer of 209,904.9 joules of heat energy to make .069 kilograms of ice at -5 degree Celsius into steam at 110 degrees Celsius. Disclaimer These videos are intended for educational purposes only (students trying to pass a class) If you design or build something based off of these videos you do so at your own risk. I am not a professional engineer and this should not be considered engineering advice. Consult an engineer if you feel you may put someone at risk.
We have a beaker filled with .069 kilograms of ice We want to find out how much heat energy in joules it will take to turn the ice with intial temperature of -5 degrees Celsius into steam at 110 degrees Celsius For the first step of this problem we need to find out how much heat it takes to raise the .069 kilograms of ice to the freezing point of zero degress Celsius. The equation for finding the heat energy needed to raise the temperature of ice is specific heat times mass of ice times the change in temperature. We can plug in our specific heat or how much energy it takes to raise 1 kilogram of ice 1 degree Celsius into the equation with the mass of ice in kilograms with the change in temperature which is 0 degrees Celsius final minus – 5 degree celsisus intial. We get a heat added of 724.6 joules. Now the ice will have to under go a phase change from ice to liquid water. This will require energy. We will need to know what the latent heat of fusion is which is the amount of energy per kilogram of ice to change from a solid to a liquid. The latent heat of fusion is 3.33 times 10 to the 5th joules per kilogram of ice. After multiplying the latent fusion times the mass of ice we get 22,977 additional joules of heat needed to convert the ice to a liquid. For the next step we need to find out how much heat energy will be required to raise the water from zero degrees to 100 degrees. We will use the specific heat for liquid water and multiply it times the mass of the ice which is now water times the heat delta of 100 degrees. We get an additional 28,883.4 joules of heat energy needed to raise the water up to the boiling point. The water will need to change phases again this time from liquid to a gas. So we will need to know what the latent heat of vaporization is. The latent heat of vaporization of water is 22.6 times 10 to the 5th joules a kilogram. This value means that for every kilogram of water it will require 22.6 times 10 to the 5th joules of energy to convert it to a gas vapor. So we can multiply this times how many kilograms of water we have which is .069 kilograms. We get an additional 155,940 joules of heat needed to get the gas to turn into a vapor. Now that gas will have a slightly higher temperature then the boiling point. In this case it has a temperature of 110 degrees Celsius. So we need to look up the specific heat for water when it is a steam. The specific heat is 2,000 joules per kilogram of steam raised by 1 degree celsisus. So now if we multiply the specific heat times the mass of steam and the change in temperature or delta of 10 degrees Celsius we get 1,380 joules of heat needed to raise the gas vapor by 10 degrees Celsius. To finish of this video lets find the total joules of heat needed to get water from ice to steam. We can add up all of the joules of heat needed for the various temperature changes and phases changes. So after we added up q1 through 5 we get an answer of 209,904.9 joules of heat energy to make .069 kilograms of ice at -5 degree Celsius into steam at 110 degrees Celsius. Disclaimer These videos are intended for educational purposes only (students trying to pass a class) If you design or build something based off of these videos you do so at your own risk. I am not a professional engineer and this should not be considered engineering advice. Consult an engineer if you feel you may put someone at risk.