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How many joules of energy are released when 45.0 g of water cools from 18.0 oC to 7.0 oC?
energy = mass x specific heat x temperature change = 45 x 4.181 x 11 J = 2069.595 J
One kilocalorie of heat energy will raise temp of water how many degrees celsius?
Change in temperature, dT,=1/m, where m is the amount of water, in kilograms. Depends on how much kilograms of water you have. The equation to use is Q=cm(dT), where c is the specific heat capacity of water (conventional use puts it 4186 J/(kg*C degrees)), m is the mass, in kilograms, of the water you're heating, dT is the amount of degrees, in Celsius (or Kelvin), that you want to heat the water by, and Q is the amount of energy needed, in Joules. 1 kilocalorie equals 4186J, solving for change in temperature (dT), dT=Q/(cm). Substituting in Q and c gives you dT=1/m. If you do not know the mass, but only volume, m=pV, where p is the density of water (convetionally 1 *10^3 kg/m^3) and V is the volume of the amount of water you have in m^3.
What energy unit is defined as the heat required to raise on kilogram of water by one degree Celsius?
I believe it is Calorie.
How much heat energy is required to raise the temperature of 15 g of water 25 degrees Celsius?
6480 calories8LBs X 15 gallons x 54 DEGREES = 6480 CALORIES
What has more thermal energy one gram of water at 75 degrees or 100 grams of water at 75 degrees?
(75'C)x(1g) < (75'C)x(100g) .'. The second option has more thermal energy.
How much energy is needed to increase the temperature of a kilogram of water 5 degrees?
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.
What is the heat of vaporization of water in joules per kilogram?
The heat of vaporization of water is 2260 joules per kilogram.
What is the latent heat of vaporization of water in joules per kilogram?
The latent heat of vaporization of water is 2260 joules per kilogram.
Energy of?
The most common unit of energy in Biology is calories. A calorie is a unit of energy, require to raise 1 kilogram of water to 4.1868 joules.
What is the specific heat of water in joules per kilogram degree Celsius?
The specific heat of water is 4186 joules per kilogram degree Celsius.
Does it it take more energy to heat the water to 100 degrees celsius or boil it?
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.
What is the amount of energy needed to raise the temperature of 1 gram of water by 1 degrees Celsius?
The specific heat capacity of water is 4.18 Joules/gram degrees Celsius. Therefore, it would take 4.18 Joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius.
How much energy would you use to raise temperature of kg of water by 2 degrees Celsius?
The amount of energy required to raise the temperature of 1 kg of water by 1 degree Celsius is approximately 4,186 Joules. Therefore, to raise the temperature by 2 degrees Celsius, you would need about 8,372 Joules of energy.
What is the unit of measurement of energy needed to heat one kilogram of water by 2 degrees F?
It is a calorie
What is is the amount of heat needed to raise the temperature of one kilogram of water one degree centigrade.?
The specific heat capacity of water is approximately 4.18 Joules per gram per degree Celsius. To raise the temperature of one kilogram (1000 grams) of water by one degree Celsius, it would require approximately 4180 Joules of heat energy.
How much joules does it take to heat water to 100 degrees?
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.
How much heat energy would be required to bring a 15 kg block of ice from -20 degrees Celsius to steam at 120 degrees Celsius?
To bring the ice block to 0 degrees Celsius, you would need 150,000 Joules (Q = mcΔT). To melt the ice at 0 degrees Celsius, you would need 3,375,000 Joules (Q = mLf). Heating the water from 0 to 100 degrees Celsius would require 1,500,000 Joules (Q = mcΔT). Turning the water to steam at 100 degrees Celsius would need 10,500,000 Joules (Q = mLv). Finally, heating the steam to 120 degrees Celsius would require 600,000 Joules (Q = mcΔT). In total, you would need 15,125,000 Joules of heat energy.
