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How much energy is needed to convert ice to water on the moon?
To convert ice to water on the moon, energy is needed to break the hydrogen bonds holding the water molecules together in the solid ice lattice. This process requires the input of heat energy to overcome the enthalpy of fusion of water, which is approximately 334 joules per gram.
How much heat is needed to convert 38.5 g of ice at 0 C to a liquid at 55 C?
To convert ice at 0°C to liquid water at 0°C, 334 J/g of heat is needed (heat of fusion). To raise the temperature of liquid water from 0°C to 55°C, 4.18 J/g°C is required (specific heat capacity of water). The total heat required would be (mass of ice x 334 J) + (mass of ice x 4.18 J/g°C x temperature difference).
How much heat in kj is needed to boil 38.0 g of water?
To calculate the heat needed to boil 38.0 g of water, we use the formula ( q = m \times L_v ), where ( m ) is the mass of the water and ( L_v ) is the latent heat of vaporization of water (approximately 2260 J/g). First, convert the mass to kilograms: 38.0 g = 0.038 kg. Then, calculate the heat: ( q = 38.0 , \text{g} \times 2260 , \text{J/g} = 85,880 , \text{J} ) or 85.88 kJ. Thus, approximately 85.9 kJ of heat is needed to boil 38.0 g of water.
How much heat is needed to freeze 100 g of water?
The heat needed to freeze 100 g of water is 334 J/g. So, for 100 g, the total heat needed would be 334 J/g * 100 g = 33,400 J.
How much heat is needed to vaporize 7.24 mL of sweat from your skin at 25 C assume sweat is only water?
To calculate the heat needed to vaporize 7.24 mL of sweat, we first convert the volume of sweat to mass, knowing that the density of water is approximately 1 g/mL. Thus, 7.24 mL of sweat is about 7.24 grams. The heat required for vaporization can be calculated using the heat of vaporization of water, which is approximately 2260 J/g. Therefore, the heat needed is 7.24 g × 2260 J/g = 16,374.4 J, or approximately 16.4 kJ.
How much energy is needed to convert ice to water on the moon?
To convert ice to water on the moon, energy is needed to break the hydrogen bonds holding the water molecules together in the solid ice lattice. This process requires the input of heat energy to overcome the enthalpy of fusion of water, which is approximately 334 joules per gram.
How much heat is required to convert 0.3 of ice at 0 to water at the same temperature?
The heat required to convert ice at 0°C to water at 0°C is known as the latent heat of fusion. For water, this value is 334 J/g. Therefore, to convert 0.3 g of ice to water at the same temperature, the heat required is 0.3 g * 334 J/g = 100.2 Joules.
Why does water take so long to warm up?
The specific heat of water determines how much energy is needed to heat water.
How much heat is needed to convert 38.5 g of ice at 0 C to a liquid at 55 C?
To convert ice at 0°C to liquid water at 0°C, 334 J/g of heat is needed (heat of fusion). To raise the temperature of liquid water from 0°C to 55°C, 4.18 J/g°C is required (specific heat capacity of water). The total heat required would be (mass of ice x 334 J) + (mass of ice x 4.18 J/g°C x temperature difference).
How much heat energy is needed to convert 50 kg of water at 8 degrees Celsius to steam at 100 degrees Celsius?
The process involves increasing the temperature of water from 8°C to 100°C and then changing its phase to steam at 100°C. The total heat energy required can be calculated using the specific heat capacity of water and the heat of vaporization. The formula Q = mcΔT can be used to find the heat energy needed, where Q is the heat energy, m is the mass of water, c is the specific heat capacity of water, and ΔT is the temperature change.
How much heat requires one litre of water evaporates?
The needed heat is 2 258 kJ.
How much heat in kj is needed to boil 38.0 g of water?
To calculate the heat needed to boil 38.0 g of water, we use the formula ( q = m \times L_v ), where ( m ) is the mass of the water and ( L_v ) is the latent heat of vaporization of water (approximately 2260 J/g). First, convert the mass to kilograms: 38.0 g = 0.038 kg. Then, calculate the heat: ( q = 38.0 , \text{g} \times 2260 , \text{J/g} = 85,880 , \text{J} ) or 85.88 kJ. Thus, approximately 85.9 kJ of heat is needed to boil 38.0 g of water.
How much coolant aka water is needed for a nuclear power station?
It depends on how much it needs to be cooled down, or transferred. The more heat there is the more coolant (water) is needed.
How much heat is needed to freeze 100 g of water?
The heat needed to freeze 100 g of water is 334 J/g. So, for 100 g, the total heat needed would be 334 J/g * 100 g = 33,400 J.
How much heat is needed to vaporize 7.24 mL of sweat from your skin at 25 C assume sweat is only water?
To calculate the heat needed to vaporize 7.24 mL of sweat, we first convert the volume of sweat to mass, knowing that the density of water is approximately 1 g/mL. Thus, 7.24 mL of sweat is about 7.24 grams. The heat required for vaporization can be calculated using the heat of vaporization of water, which is approximately 2260 J/g. Therefore, the heat needed is 7.24 g × 2260 J/g = 16,374.4 J, or approximately 16.4 kJ.
How much energy is needed to melt 0.25 moles of water?
The necessary heat is 9,22 joules.
How much heat is needed to raise 8 kg of water from 4 degrees Celsius to 83 degrees Celsius?
The heat needed can be calculated using the formula: Q = mc∆T, where Q is the heat, m is the mass of water, c is the specific heat capacity of water (4186 J/kg°C), and ∆T is the change in temperature. Plug in the values to find the heat needed.
