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The energy is 103,6 kcal.
6.3 L
A change of 100 degrees K is equal to a change of 100 degrees Celsius.
dT = (i)(Kf)(m) is the equation that can be used to model the freezing point depression. dT represents the change change in temperature. i represents the amount of ions formed from the dissolution of the solute. Kf represents the constant of the freezing point depression of water. m is the molality of the solute in solution. dT = (3)(1.86 degC/m)(2.65m) dT = 14.8 degC 0-14.8 degC = -14.8 degC So the freezing point is -14.8 degrees C. The reason why this value was subtracted from zero is because the presence of a solute lowers the freezing point.
The molar volume (22,414 L at oC and 1 at) is a constant at a given temperature.
one calorie of heat is able to raise one gram of water one degree Celsius so 400 calories could raise 1g of water 400 degrees, so it would raise the 80g by(400/80) 5 degrees Celsius plus the initial temp of 10 degrees, the 80g of water would have a final temp of 15 degrees Celsius
The energy is 103,6 kcal.
The needed heat is:Q = 10 x 20 x 0,031 = 6,2 calories
A PWR has an inlet water temperature of 275 degC and outlet 325 degC
It would be approx 9042 litres.
Water is transformed in vapors.
(4.184 J/g*degC)(400g)(40.0*degC-80.0*degC)+(200g)
40
In tokamak reactors, approx 300 million degC
The new volume is 544,5 l.
178.9 Celsius is 354 Fahrenheit. Multiply by nine fifths and add thirty-two.
Degrees Celsius. Absolute (SI units) are Kelvin=degC+273.15