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.
To calculate the heat absorbed by the water, you can use the formula: heat = mass * specific heat capacity * temperature change. First, determine the specific heat capacity of water (4.18 J/g°C). Then, plug in the values: heat = 15 g * 4.18 J/g°C * 3.0°C. The heat absorbed by the water is 188.1 Joules.
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.
You would need 20,920 Joules of heat to raise the temperature of 1kg of water by 5°C. This value is calculated using the specific heat capacity of water, which is 4186 J/kg°C.
The specific heat capacity of water is 4.18 J/g°C. To calculate the energy required to raise 21 kg of water by 2 degrees Celsius, use the formula: Energy = mass x specific heat capacity x temperature change. Plugging in the values, the energy required is 21,084 Joules.
q (heat energy in Joules) = mass * specific heat * change in temp 1st problem: q = (100 g H2O)(4.180 J/gC)(100 C - 50 C) = 20900 Joules ---------------------- 2nd problem: q = (100 g H2O)(4.180 J/gC)(70 C - 60 C) = 4180 Joules --------------------- As you can see from 50 C to 100 C takes much more heat energy as one would intuitively think, 20900 J/4180 J = 5 times as much energy.
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The latent heat of condensation of steam is 2260 Joules per gram (539.3 cals/g). So the amount of heat released by 12.4 g = 12.4*2260 Joules = 28,024 Joules or 6687 cals.
70 calories per gram. (The specific heat capacity of water is 1 calorie per gram per degree C.) This could be converted into Joules if necessary using the conversion factor of 1 calorie = 4.18400 Joules.
The heat lost by water at 0 degrees Celsius to change to ice is equal to the heat of fusion of water, which is about 334 joules per gram. So, for 2 grams of water, the heat loss would be 2 * 334 = 668 joules.
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.
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.
To calculate the heat absorbed by the water, you can use the formula: heat = mass * specific heat capacity * temperature change. First, determine the specific heat capacity of water (4.18 J/g°C). Then, plug in the values: heat = 15 g * 4.18 J/g°C * 3.0°C. The heat absorbed by the water is 188.1 Joules.
Heat of vaporization of water is 2.26 x 106 joules per kg. Therefore 1 gram of water will need 2.26 x 103 joules.
q( in Joules ) = mass * specific heat * change in temperature [ convert temps--Tf = Tc(1.80) + 32 ] q = (40 g)(0.90 J/gC)(61.1o C - 22.8o C) = 1.4 X 103 Joules =============
Measured in joules, the process would take 334,584 J to occur completely.
The necessary heat is 9,22 joules.
q = (250 g)(0.46 J/gC)(300 C - 27 C) = 3.1 X 104 Joules -------------------------