Answer to Question #76245 in Molecular Physics | Thermodynamics for Prashant

Question

2 kg of water is heated from 0°C to 100°C and converted into steam at the same
temperature. Calculate the increase in entropy, given that specific heat of water is
4.18 × 103
J kg−1
K−1
and Latent heat of vaporisation is 2.27 × 107
J kg−1
.

Answer 1

1)
T_1=0+273=273 K.
T_2=100+273=373 K.
ΔS_1=mc ln⁡〖T_2/T_1 〗=(2)(4180) ln⁡〖373/273〗=2609 J/K
2)
ΔS_2=Q/T
Q=mL.
ΔS_2=mL/T=((2)(2.27∙ 〖10〗^7))/373=121716 J/K
3) The increase in entropy:
ΔS=ΔS_1+ΔS_2=2609+121716=124325 J/K=124 kJ/K.