# Converting kJ provided to a closed system into final temperature of the system

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.

• I think what you are missing is volume or pressure. Sep 23 '21 at 18:11
• 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 &deg; for degrees symbol, &mu;, &Omega; &omega;, etc. Sep 23 '21 at 18:51
• Did you forget the latent heat of evaporation when converting the water to steam? Sep 23 '21 at 18:53