# 1kg of water at 0 degree celsius is fully converted into steam at 100 degree celsius at normal pressure.Calculate the change in entropy.The specific heat capacity of water is 4.18*10^3 J kg^-1 K^-1...

1kg of water at 0 degree celsius is fully converted into steam at 100 degree celsius at normal pressure.Calculate the change in entropy.The specific heat capacity of water is 4.18*10^3 J kg^-1 K^-1 and latent heat of vaporisation is 2.24*10^6 J kg^-1

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Entropy is the reversible enthalpy change of a process divided by T. Assuming the heat changes in the given situations to be reversible in nature:

DeltaS = Delta(S_(heating)) + Delta(S_(evaporation))

= Integral (dq_(rev) / T) + (q_2)/T_steam

For heating up of water, dq = m s dT, (where s is the specific heat of water)

Thus,

Delta(S_(heating)) = integral ms dT / T = msln(T2/T1)

(q2) = m L , with L the latent heat of vaporisation.

Therefore,

Delta(S) = m s ln((T_2)/(T_1)) + m L / T_2

= 1kg* 4.18*10^3J/(kg K) * ln(373/273) + 1kg*2.24*10^6 J/kg / (373K)

= 7.31*10^3 J/K

=7.31 kJ/K