Question

An adiabatic tank contains 120 kg of liquid water initially at 70 °C. The water within the tank is utilized to quench (cool) a 15 kg steel block initially at 300 °C. The steel block is dropped into the tank, and after some time the steel block and the water in the tank reach thermal equilibrium.

Determine the total entropy change during the quenching process.

Specific heat of steel : C_steel= 0.50 kJ/kg*K.

Specific heat of water : C_water= 4.18 kJ/kg*K

Answer 1

Let the equilibrium temperature reached be Teq

heat lost by steel=heat gained by water (heat =Q=mass * sp.heat* change in temperature)

15 kg*0.50 KJ/kg k*(Teq-300)=120kg *4.18 KJ/kg K*(Teq-70)

                              (Teq-300K)=66.88(Teq-70K)

                               Teq=66.51K

Sh=entropy change of hot body= mass Csteel ln T/Th=15 kg* 0.50 KJ/kg/K ln ( 66.51/300)

                                                                                     = 7.5 KJ/K (-1.51)=-11.32 Kj/K

Sc=entropy change of cold body= mass Cwater ln T/Tc =120 Kg*4.18 KJ/kg K ln(66.51/70)

                                                                                 =(501.6)(-0.051)=-25.58Kj/K

total entropy change=Sh+Sc=-11.32-25.58=-36.9kj/k