Question

An isolated piston-cylinder device contains 5 L of saturated liquid water at a constant pressure of 150 kPa. An electric resistance heater inside the cylinder is now turned on, and 2200 kJ of energy is transferred to the steam. Determine the entropy change of the water during this process, in kJ/K.

Answer 1

entropy change = 5.72 KJ/K

Step-by-step explanation:

The energy balance for the system is given as;

ΔU = W_in - W_out

However, the entropy change of water is given as;

ΔS = m(S2 - S1)

ΔS = (V1/α1)*((h2 -h_liq150)/h_evap150)*(s_evap150)

This simplifies to;

ΔS = (V1/α1)*((h_liq150 + W_in*α1/V1 - h_liq150)*(s_evap150/h_evap150)

ΔS = W_in*(s_evap150/h_evap150)

We are given a constant pressure of 150 kPa

From saturated water table attached, at 150 KPa,

h_evap150 = h_fg = 2226 KJ/Kg

s_evap150 = s_fg = 5.7894 KJ/Kg.K

Since we are given W_in as 2200 kJ.

Thus;

ΔS = 2200(5.7894/2226)

ΔS = 5.72 KJ/K