Wiki User
∙ 7y agoApprox 2940 Joules.
Wiki User
∙ 6y agoTo 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 convert 50 kg of water at 80 degrees Celsius to steam at 100 degrees Celsius, you need to calculate the heat energy needed to raise the temperature of water from 80 to 100 degrees Celsius (specific heat capacity of water) and then the heat energy needed for water to steam phase change (latent heat of vaporization of water). The total heat energy required can be calculated using the formula: Q = mcΔT + mL, where Q is the heat energy, m is the mass, c is the specific heat capacity, ΔT is the temperature change, and L is the latent heat of vaporization.
The correct value for the specific heat of water is 4.1868 joules per gram per degree Celsius. This means that it takes 4.1868 joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius.
This value is 22,418 kJ.
steam scalds are more serious than boiling water scalds because steam when condenses onto the more cooler skin, it loses latent heat of vaporization (to become water at 100 degrees Celsius) also it loses thermal capacity to become equal to the temperature of the skin (37 degrees Celsius). boiling water loses only thermal capacity as it cools down to 37 degrees Celsius from 100 degree 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.
When steam at 100 degrees Celsius condenses, it releases 2260 Joules of energy per gram. Therefore, for 1000g of steam, the heat released would be 2,260,000 Joules (2260 J/g * 1000 g).
The specific heat capacity of steam at 100 degrees Celsius is approximately 2.08 J/g°C. This means it takes 2.08 joules of energy to raise the temperature of 1 gram of steam by 1 degree Celsius at that temperature.
You mean how much heat energy will be lost/transferred as you are losing Joules here. All in steam, so a simple q problem and no change of state. 2.67 kg = 2670 grams q = (2670 grams steam)(2.0 J/gC)(105 C - 282 C) = - 9.45 X 105 Joules ----------------------------------- This much heat energy must be lost to lower the temperature of the steam.
To calculate the energy required to heat the steam, you need to use the specific heat capacity of steam, which is 2.03 J/g°C. The change in temperature is 100.0 - 29.25 = 70.75°C. The energy required can be calculated using the formula: energy = mass * specific heat capacity * temperature change. Substituting the values gives: energy = 25.0 g * 2.03 J/g°C * 70.75°C = 3608.375 J.
To convert 50 kg of water at 80 degrees Celsius to steam at 100 degrees Celsius, you need to calculate the heat energy needed to raise the temperature of water from 80 to 100 degrees Celsius (specific heat capacity of water) and then the heat energy needed for water to steam phase change (latent heat of vaporization of water). The total heat energy required can be calculated using the formula: Q = mcΔT + mL, where Q is the heat energy, m is the mass, c is the specific heat capacity, ΔT is the temperature change, and L is the latent heat of vaporization.
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.
To vaporize 1 gram of boiling water at 100 degrees Celsius, it would require approximately 2260 joules, which is equivalent to about 0.54 calories. This energy is needed to break the intermolecular bonds holding water molecules together, allowing them to escape into the gas phase.
Steam has more heat energy compared to water at 100 degrees Celsius because steam is in a gaseous state and has absorbed additional heat energy to transition from a liquid to a gas.
The correct value for the specific heat of water is 4.1868 joules per gram per degree Celsius. This means that it takes 4.1868 joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius.
Steam at 100 degrees Celsius contains more heat energy compared to liquid water at the same temperature. When steam comes into contact with skin, it releases this extra heat quickly, causing more severe burns compared to liquid water of the same temperature.
To convert 12.5 grams of ice at 0 degrees Celsius to steam at 100 degrees Celsius, you would need to provide heat energy for three main processes: heating the ice from 0 degrees Celsius to 100 degrees Celsius, melting the ice into water at 0 degrees Celsius, and then heating the water from 0 degrees Celsius to steam at 100 degrees Celsius. The total calorie requirement would be determined by the specific heat capacities and heat of fusion and vaporization of water.