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The energy content of water at 373 K can be calculated using the specific heat capacity of water, which is 4.18 J/g°C. At 373 K, the specific heat capacity would be approximately 1.00 kcal/kg°C. Therefore, the energy content of 1 kg of water at 373 K would be approximately 373 kcal.
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How much joules of energy is present in 1 kg of water?
The amount of energy present in 1 kg of water is typically around 4,186,000 joules at room temperature. This value represents the energy required to raise the temperature of 1 kg of water by 1 degree Celsius.
How much energy would you use to raise temperature of kg of water by 2 degrees Celsius?
The amount of energy required to raise the temperature of 1 kg of water by 1 degree Celsius is approximately 4,186 Joules. Therefore, to raise the temperature by 2 degrees Celsius, you would need about 8,372 Joules of energy.
How much energy will it take to raise 7.3 kg of water from 10 degrees Celsius to 90 degrees celsius?
To find the energy required to raise the temperature of water, you can use the formula: Energy = mass x specific heat capacity x temperature change. The specific heat capacity of water is 4.18 J/g°C. Therefore, the energy required to raise 7.3 kg of water from 10°C to 90°C would be 7.3 kg x 4.18 J/g°C x (90°C - 10°C).
How much heat is contained in 100 kg of water at 60.0 C?
The amount of heat contained in 100 kg of water at 60.0°C can be calculated using the formula Q = mcΔT, where Q is the heat energy, m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature. Given that the specific heat capacity of water is 4186 J/kg°C, the heat contained in 100 kg of water at 60.0°C would be Q = 100 kg × 4186 J/kg°C × (60.0°C - 20.0°C).
How much energy is necessary to heat 1.0 kg of water from room temperature (20 and deg C) to its boiling point (Assume no energy loss.) (b) If electrical energy were used how much would it cost at?
The energy needed to heat 1.0 kg of water from 20°C to 100°C is 334 kJ (specific heat capacity of water is 4.18 kJ/kg°C). To calculate the cost, you would need to know the cost of electricity per kilowatt-hour. If, for example, the cost is $0.12 per kWh, the cost to heat this water would be around $0.04 (334 kJ = 0.093 kWh, and 0.093 kWh x $0.12/kWh = $0.011).
Which contain more heat 1 kg of steam at 373k or 1 kg of water at 373k?
1 kg of steam at 373 K contains more heat than 1 kg of water at 373 K because steam has a higher specific heat capacity and latent heat of vaporization than water. This means more heat energy is required to convert water at 373 K into steam at 373 K.
373 grams is how many kilogram?
A kilogram (kg) is 1000 grams, therefore 373 grams is .373 kg.
What is 373 grams in kilograms?
kilo = 1000kilogram = 1000g373g = 0.373kg
How much joules of energy is present in 1 kg of water?
The amount of energy present in 1 kg of water is typically around 4,186,000 joules at room temperature. This value represents the energy required to raise the temperature of 1 kg of water by 1 degree Celsius.
How do you convert 373 g into kg?
Divide by 1000. So 373 / 1000 = 0.373 kilograms.
How many kilograms are in 373 grams?
373 g = 0.373 kgTO convert from g to kg, divide by 1000.
How many Kg are in 373g?
373 grams = 0.373 kilograms.
How much energy is needed to raise the temperature of 2kg of water by 1ᵒC?
To raise the temperature of 2 kg of water by 1°C, you need 8,400 joules of energy. This is calculated using the specific heat capacity of water, which is approximately 4,200 joules per kilogram per degree Celsius (J/kg°C). Therefore, the energy required is 2 kg × 4,200 J/kg°C × 1°C = 8,400 J.
How much energy is required to make 1 kg of steam from 1 ltr of water?
The answer will depend on the starting temperature of the water. It will also depend on the pressure.
How much energy was transferred to the water of 1.3 kg of water equals from 20 degrees to 100 degrees c?
1.3 kg water = 1300 grams. q(Joules-heat energy) = mass * specific heat * change in temperature q = (1300 g)(4.180 J/gC)(100 C - 20 C) = 4.3 X 105 Joules of heat energy ========================
A person has a pumpkin that has a mass of 4.8 kilograms If 0.9 of its mass is water how much of its mass is water and how much of its mass is not water?
The pumpkin's mass is 4.8 kg, and 0.9 of its mass is water. This means 0.1 of its mass is not water. To find the mass of water: 4.8 kg x 0.9 = 4.32 kg of water To find the mass of not water: 4.8 kg - 4.32 kg = 0.48 kg of not water
how much energy is needed to increase the temperature of 1kg of water from 20°C to 40°C. (specific heat capacity of water is 4kJ/°C/kg)?
ewan
