If we double the thermal energy of a cup of boiling water100c, will the new temperature will be near 200C If we double the thermal energy of a cup of boiling water the new temperature will be near 200 0C
kindly can someone explain to me the relvent formula for thermal energy and temperature
 A: If you try to put more thermal energy into a cup of water which is already at 100C, the water will begin to boil into steam at 100C, and the escaping steam will carry away all the extra energy you are putting into the water. As long as there remains some water in that cup, its temperature will be stuck at (or very close to) 100C.
But if you have some way to dump the extra heat into the water extremely rapidly, then the normal boiling process can be avoided- thereby superheating the water- for a very brief time.
For example, if you raise the water's temperature at a rate of, say, 100C per microsecond, you can get the water all the way up to about 260C, but it will then explode into steam after no more than about a quarter of a microsecond.
The temperature at which the boiling explosion time delay goes to zero is called the thermodynamic limit of superheat, which for pure water at atmospheric pressure is about 300C.
A: The case of a cup of water is complicated by the phase change from liquid to gas and the non-negligible interactions between water molecules. Let us instead consider the much simpler scenario of a monoatomic ideal gas.
The thermal energy is simply the total kinetic energy of all particles, which amounts to $E=\frac 3 2 kNT$ by the equipartition theorem. The thermal energy is indeed proportional to temperature - however, we have to be careful to substitute absolute temperature, i.e. in Kelvin.
As 100 °C is 373.15 K, doubling the thermal energy of a 100 °C ideal gas will result in a temperature of 746.3 K = 473.15 °C. Much larger than the 200 °C you have proposed.
