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How much heat is necessary vaporize 500 g of water at ts boiling point?
To calculate the heat required to vaporize 500 g of water at its boiling point, you can use the formula ( Q = m \times L_v ), where ( Q ) is the heat energy, ( m ) is the mass of the water, and ( L_v ) is the latent heat of vaporization. The latent heat of vaporization for water is approximately 2260 J/g. Therefore, the heat required is ( Q = 500 , \text{g} \times 2260 , \text{J/g} = 1,130,000 , \text{J} ) or 1.13 MJ.
How long to heat 500 liter water with 3kw element?
Specific Heat of Water = 4.1813 kJ/l/K If starting temperature is say 10°C - 500 Litres * 4.1813 (100°C - 10°C) = 188158.5 kJ 188158.5kJ / 3kW = 62719.5 Seconds = 17.42 Hours
How much heat must be absorbed by a 500 g pot of water in order to raise the temperature from 20c to 30c?
To calculate the heat absorbed by the water, you can use the formula ( Q = mc\Delta T ), where ( Q ) is the heat absorbed, ( m ) is the mass of the water, ( c ) is the specific heat capacity of water (approximately 4.18 J/g°C), and ( \Delta T ) is the change in temperature. For a 500 g pot of water raising the temperature from 20°C to 30°C (( \Delta T = 10°C )), the calculation would be: [ Q = 500 , \text{g} \times 4.18 , \text{J/g°C} \times 10 , \text{°C} = 20,900 , \text{J} ] Thus, 20,900 joules of heat must be absorbed.
Which requires more energy converting 50 grams of ice at 0C to 50 grams of water at 100C or converting 50 grams of water at 100C to 50 grams of steam at 100C?
would require more heartmelting 0 c ice turning100 water into steam These terms only appear in links pointing to this page: 500 g500 g c
How much heat is absorbed when 500. g of water Cp 4.184 Jg C goes from 25.0 C to 35.0 C?
To calculate the heat absorbed by the water, use the formula ( q = m \cdot C_p \cdot \Delta T ), where ( q ) is the heat absorbed, ( m ) is the mass of the water (500 g), ( C_p ) is the specific heat capacity (4.184 J/g°C), and ( \Delta T ) is the change in temperature (35.0°C - 25.0°C = 10.0°C). Plugging in the values: [ q = 500 , \text{g} \times 4.184 , \text{J/g°C} \times 10.0 , \text{°C} = 20920 , \text{J} ] Therefore, the heat absorbed is 20,920 J.
How do you increase the the temperature of 500 grams of water from 20 degrees Celsius to 100 degrees Celsius?
q(joules) = mass * specific heat * change in temperature q = (500 grams H2O)(4.180 J/goC)(100o C - 20o C) = 1.7 X 105 joules ================add this much heat energy to the water
How much heat is necessary vaporize 500 g of water at ts boiling point?
To calculate the heat required to vaporize 500 g of water at its boiling point, you can use the formula ( Q = m \times L_v ), where ( Q ) is the heat energy, ( m ) is the mass of the water, and ( L_v ) is the latent heat of vaporization. The latent heat of vaporization for water is approximately 2260 J/g. Therefore, the heat required is ( Q = 500 , \text{g} \times 2260 , \text{J/g} = 1,130,000 , \text{J} ) or 1.13 MJ.
How long will it take a 500 W heater to raise the temperature of 400g of water from 15.0 C to 98.0 C?
Using the formula Q = mcΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity of water, and ΔT is the temperature change gives you 70,800 J of energy needed. With a 500W heater, it will take 141.6 seconds or approximately 2.36 minutes to heat the water.
In terms of particles Why does an iceberg have more heat energy than molten iron?
This is based on mass and heat capacity. an iceberg with a mass of 500 metric tons would have a heat content of ~ 35.6 million kilo-calories. 1 kg of molten iron would have a heat content of 260.372 kilo-calories (500 metric tons of molten iron has a heat content of 130,186 million kilo-calories)
How much energy would be required to melt 500.g of ice at 0.00 degrees * C to water at 0.00 degrees * C ?
200
How much heat is necessary to vaporize 500 grams of ice at its freezing point?
The heat required to vaporize 500 grams of ice at its freezing point is the sum of the heat required to raise the temperature of the ice to its melting point, the heat of fusion to melt the ice, the heat required to raise the temperature of water to its boiling point, and finally the heat of vaporization to vaporize the water. The specific heat capacity of ice, heat of fusion of ice, specific heat capacity of water, and heat of vaporization of water are all needed to perform the calculations.
If a organism consumes 500 calories of energy thn how much is it lost for heat?
The exact details will surely vary from one organism to another, but in general, most of the energy is eventually converted to heat. A small amount may be stored as an energy reserve, for example as sugar, or as fatty tissue.
How much energy is needed to heat 0.5kg of Lead?
The specific heat (heat capacity) of lead is 26,65 J/mol.K; 64,31 J for heating 500 g with 1 K.
How much heat is absorbed when 500. g of water Cp 4.184 JgoC goes from 25.0 oC to 35.0 oC?
The heat absorbed by water can be calculated using the formula: Q = m * Cp * ΔT, where Q is the heat absorbed, m is the mass of the water (500 g), Cp is the specific heat capacity of water (4.184 J/g°C), and ΔT is the change in temperature (35.0°C - 25.0°C = 10.0°C). Plugging these values in, we get: Q = 500 g * 4.184 J/g°C * 10.0°C = 20,920 J. So, the heat absorbed by the water is 20,920 Joules.
How A 500 g piece of iron changes 7 C when heat is added. How much heat energy in joules is produced due to this change in temperature?
To calculate the heat energy produced, you need to use the formula Q = mcΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity of iron (0.45 J/g°C), and ΔT is the change in temperature. Plugging in the values, you get Q = (0.45 J/g°C)(500 g)(7°C) = 1575 Joules.
How long to heat 500 liter water with 3kw element?
Specific Heat of Water = 4.1813 kJ/l/K If starting temperature is say 10°C - 500 Litres * 4.1813 (100°C - 10°C) = 188158.5 kJ 188158.5kJ / 3kW = 62719.5 Seconds = 17.42 Hours
If 500 J of energy is put in a computer but 120J is wasted as heat what is the efficiency?
You have to divide the useful energy by the input energy, in this case, 380 Joule / 500 Joule.
