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Calculate the amount of energy (in kJ) necessary to convert 497 g of liquid water from 0°C to water vapor at 167°C. The molar heat of vaporization (Hvap) of water is 40.79 kJ/mol. The specific heat for water is 4.184 J/g · °C, and for steam is 1.99 J/g · °C. (Assume that the specific heat values do not change over the range of temperatures in the problem.)

Respuesta :

Answer:

The amount of energy required will be 1400.61 KJ.

Explanation:

Mass of water liquid m= 497 g

Mole ,n= m/18

n=497/18 mole

n=27.6 mole

Total heat=Latent heat of water at constant temperature 0°C + Sensible heat of water from temperature 0 °C to 100° C+Sensible heat of steam from 100 °C to 167°C.

We know that

Sensible heating =m Cp ΔT

So the total heat ,Q

Q= 27.6 x 40.79 x 1000 + 497 x 4.184 x 100+ 497 x 1.99 x 67   J

Q=1400.61 KJ

So the amount of energy required will be 1400.61 KJ.

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