Want this question answered?
2kj
The necessary heat is 0,155 kJ.
The energy is 103,6 kcal.
To answer this, you need to know the ∆Hfusion of water, which happens to be 334 J/g. So, to melt 12.8 g of ice at 0ºC, the joules needed = (12.8 g)(334 J/g) = 4275 joules
Ice (solid water) - the freezing point of water is o 0C; but because of the sublimation also gaseous molecules exist.
10,267 kJ are needed
2kj
It boils
Water is transformed in vapors.
A PWR has an inlet water temperature of 275 degC and outlet 325 degC
The necessary heat is 0,155 kJ.
In tokamak reactors, approx 300 million degC
The solubility increase from 38,7 g KCl/100g water to 40,7 g KCl/100 g water.
It is in the gas phase.
(4.184 J/g*degC)(400g)(40.0*degC-80.0*degC)+(200g)
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
The metric unit for heat is the calorie - the heat required to raise 1 gram of water by 1 deg C. In the SI unit system it would be the kilocalorie - the heat to raise 1 kg by 1 degC