2,26 Kj are necessary
To calculate the energy needed to boil 90 grams of water, we use the formula ( q = m \cdot L ), where ( q ) is the heat energy, ( m ) is the mass of the water, and ( L ) is the latent heat of vaporization for water, which is approximately 2260 kJ/kg. For 90 grams (0.09 kg), the calculation is ( q = 0.09 , \text{kg} \cdot 2260 , \text{kJ/kg} ), resulting in about 203.4 kJ of energy needed to boil the water.
63 g of water are needed.
This value is 22,418 kJ.
Approx 2940 Joules.
The answer is 3,211 g.
9.1x10^2J
The heat of vaporization of water is 40.79 kJ/mol. First, determine the number of moles in 24.40 grams of water. Then, convert moles to joules using the molar heat of vaporization. This will give you the amount of heat needed to vaporize 24.40 grams of water.
Voltage is electrical pressure and Joules is a power rating so other info is needed
0000000000000000.11 joules
46 calories (or 192, 464 joules) for each Celsius degree.
It takes 4.184 joules of energy to change the temperature of 1 gram of water by 1 degree Celsius.
0.0796
A lot
63 g of water are needed.
This value is 22,418 kJ.
q(joules) = mass * specific heat * change in temperature ( 8 kg = 8000 grams ) q = (8000 grams H2O)(4.180 J/gC)(70o C - 20o C) = 1.7 X 106 joules ============
419.1 Joules are required to heat one gram of liquid water from 0.01 degC to 100 deg C. So the answer is 419.1*46 = 19278.6