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If 20.6 grams of ice at -35c are heated and vaporized to all gas at 110c how much energy is required?
To calculate the energy required to heat and vaporize the ice, you need to consider the heat needed for each step:
- Heat the ice from -35°C to 0°C (specific heat of ice).
- Melt the ice at 0°C (heat of fusion).
- Heat the water at 0°C to 100°C (specific heat of water).
- Vaporize the water at 100°C (heat of vaporization).
- Heat the steam from 100°C to 110°C (specific heat of steam). Adding all these energies together will give you the total energy required.
What else can I help you with?
How much energy in kilocalories is absorbed when 9.0 mol of water is vaporized?
Molecular weight of water is 18 so 9 moles is 162 grams. Latent heat of vaporisation is 550 cal/gram, so the amount of heat is 89100 cal or 89.1 kcal.
How many KJ of energy is in 100g of chicken?
1 kilogram = 1000 grams. You now have all the information required to work out, for yourself, the answer to this and similar questions. And whether that is chicken or not makes no difference.
How much energy is required to raise the temperature of 5 grams of iron from -10ºC to 30ºC?
The specific heat capacity of iron is 0.45 J/g°C. To calculate the energy required, you can use the formula: Energy = mass x specific heat capacity x change in temperature. Plugging in the values, Energy = 5g x 0.45 J/g°C x (30°C - (-10°C)). This calculation would give you the energy in joules required to raise the temperature of 5 grams of iron from -10ºC to 30ºC.
How much energy is gained by 3 grams of ice when it changes to water?
The energy needed to change ice into water is called the heat of fusion. For ice, this value is around 334 joules per gram. So, for 3 grams of ice, the energy gained when it changes to water would be around 1002 joules (334 joules/gram * 3 grams).
The specific heat of gold is 0.131 Joules per gram. Celsius How much energy is required to heat 1.3 grams of gold from 25 Celsius to 46 Celsius?
The change in temperature is 21 degrees Celsius. To calculate the energy required, we use the formula: Energy = mass * specific heat * change in temperature. Plugging in the values, Energy = 1.3g * 0.131 J/g°C * 21°C = 35.247 Joules. Therefore, 35.247 Joules of energy is required to heat 1.3 grams of gold from 25°C to 46°C.
How is stoichiometry used to calculate energy absorbed when a mass liquid boils?
Grams liquid × mol/g × Hvap
How many kilojoules of heat are absorbed when 70.00 grams of water is completely vaporized at its boiling point?
158.1Kj
How much energy in kilocalories is absorbed when 9.0 mol of water is vaporized?
Molecular weight of water is 18 so 9 moles is 162 grams. Latent heat of vaporisation is 550 cal/gram, so the amount of heat is 89100 cal or 89.1 kcal.
If a 150 gram iron bar is heated to 250c a second iron bar with a mass or 300 grams is heated to 250c which bar has the most thermal energy?
The second iron bar with a mass of 300 grams has the most thermal energy because thermal energy is proportional to both mass and temperature. The greater mass of the second iron bar means it will have more thermal energy compared to the first iron bar with a mass of 150 grams, even if they are both at the same temperature of 250°C.
How many KJ of energy is in 100g of chicken?
1 kilogram = 1000 grams. You now have all the information required to work out, for yourself, the answer to this and similar questions. And whether that is chicken or not makes no difference.
How much energy is required to raise the temperature of 5 grams of iron from -10ºC to 30ºC?
The specific heat capacity of iron is 0.45 J/g°C. To calculate the energy required, you can use the formula: Energy = mass x specific heat capacity x change in temperature. Plugging in the values, Energy = 5g x 0.45 J/g°C x (30°C - (-10°C)). This calculation would give you the energy in joules required to raise the temperature of 5 grams of iron from -10ºC to 30ºC.
How much energy is required to evaporate 5 grams toluene?
Toluene had heat of vaporization of 38.06 kJ/mol Toluene molecular mass is 92.14 g/mol You just need to convert heat of vaporization into kJ/g basis and multiply your 5 gram to get the required amount of heat.
How is it possible that something can have a high temperature but very little thermal energy?
Thermal energy is a product of two variables; the temperature, and the mass. If two objects having the same mass were heated to the same temperature, they would have the same thermal energy. If an object weighing ten grams was heated to 1000º C, it would have less thermal energy than an object weighing 2 tons, heated to 100º C. To demonstrate this, imagine the amounts of ice each of the above objects could melt.
How much energy must be absorbed by 150 grams of ice?
To get ice to its melting point, it must absorb 334 joules of energy per gram of ice. So, for 150 grams of ice, the total energy required would be 50,100 joules (334 J/g * 150 g).
How much energy is gained by 3 grams of ice when it changes to water?
The energy needed to change ice into water is called the heat of fusion. For ice, this value is around 334 joules per gram. So, for 3 grams of ice, the energy gained when it changes to water would be around 1002 joules (334 joules/gram * 3 grams).
The specific heat of gold is 0.131 Joules per gram. Celsius How much energy is required to heat 1.3 grams of gold from 25 Celsius to 46 Celsius?
The change in temperature is 21 degrees Celsius. To calculate the energy required, we use the formula: Energy = mass * specific heat * change in temperature. Plugging in the values, Energy = 1.3g * 0.131 J/g°C * 21°C = 35.247 Joules. Therefore, 35.247 Joules of energy is required to heat 1.3 grams of gold from 25°C to 46°C.
How many joules of energy are necessary to melt 46.0 grams of ice?
To calculate the energy required to melt ice, we use the formula ( Q = m \cdot L_f ), where ( Q ) is the heat energy, ( m ) is the mass of the ice, and ( L_f ) is the latent heat of fusion for ice, approximately 334 joules per gram. For 46.0 grams of ice, the energy required would be ( Q = 46.0 , \text{g} \times 334 , \text{J/g} ), which equals about 15,364 joules. Therefore, approximately 15,364 joules are necessary to melt 46.0 grams of ice.
