I'm working a closed system in which starts with 500 grams of steam and 500 grams of water at a temperature of 100 Celsius; 2199.26 kJ is put into the system. I'm trying to find the final temperature of the system.

Using $$Q(_1-_2) = E_2-E_1 $$ I set $E_1$ equal to 0 since it's the initial state resulting in $E_2$ = $Q(_1-_2)$ which $E_2 = 2199.26 *kJ $

Using the assumption the rest of the water is turned to steam with the increased temperature, results in 1 kg of steam.

To convert the kinetic energy to temperature, dividing the energy by the mass $$ \frac{2199.26* kJ}{1* kg} = \frac {2199* J}{1* g} $$

using the specific heat of steam $2.03 \frac{J}{gC}$ results in $$ \frac{2199* JgC}{2.03* Jg} = 1083.37*℃ $$

That seems a bit high while thinking logically and was wondering if I'm missing a step.

  • $\begingroup$ I think what you are missing is volume or pressure. $\endgroup$
    – NMech
    Sep 23 '21 at 18:11
  • $\begingroup$ Tip: 'KJ' = kelvin-joules. I think you mean 'kJ' for kilojoules. Similarly 'Kg' would be kelvin-grams. SI standard recommends space between numerals and units and convention is that units are not italicised to differentiate them from variables. SE supports HTML entities too so you can use ° for degrees symbol, μ, Ω ω, etc. $\endgroup$
    – Transistor
    Sep 23 '21 at 18:51
  • 1
    $\begingroup$ Did you forget the latent heat of evaporation when converting the water to steam? $\endgroup$
    – Transistor
    Sep 23 '21 at 18:53
  1. We need the pressure of the mixture to know the latent heat of vaporization, but since we are using 100 C, we'll assume it's at 1 atmosphere.

  2. Convert the liquid to steam using latent heat of vaporization. If your energy input isn't enough to turn it all into vapor, the final temp is 100 C. If you have energy left over it will go into the next step.

  3. Using the specific heat capacity of steam vapor, calculate the temperature rise of the all-vapor steam from the remaining energy.


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