Question

A 1.00Kg horseshoe at 900°C is put in a 4.00Kg bath of water at 10°C. assuming that no energy is lost by heat to the surroundings, determine the total entropy change of the wather + horesehoe system. (HINT think about dQ_r)

Answer 1

Q1+Q2 =0 - no energy is lost by heat to the surroundings

Q1=C1*M1*(T - T1) Q2=C2*M2*(T - T2)

C1*M1*(T - T1) + C2*M2*(T - T2) = 0
C1*M1*T - C1*M1*T1 + C2*M2*T - C2*M2*T2 = 0

T= (C1*M1*T1 + C2*M2*T2)/(C1*M1 + C2*M2)

T - result temperature
C1 = Iron Specific Heat at 10C = 0.444 (kJ/kgK)
M1 = horse shoe mass = 1.0 (kg)
T1 = temperature 900 C = 1173.15 K

C2 = Water Specific Heat at 10C = 4.186 (kJ/kgK)
M2 = water mass = 4.0 (kg)
T2 = water temperature 10 C = 283.15 K

T = (0.444*1*1173.15 + 4.186*4*283.15)/(0.444*1 + 4.186*4) = 306.14 K
T = 32.99 C ~ 33 C
Tf = 33 C = 306.15 K
c_iron = 0.450 kJ/kg.K
c_water = 4.18 kJ/kg.K

Entropy chnage can be evaluate using Tds equation,
Tds = dh - vdp, both iron and water are incompressible dp~0
Tds = dh = cdT
∫ds = ∫(c/T) dT, integrate from initial state to final state.
∆s = c*ln(Tf/Ti)
∆S = m*c*ln(Tf/Ti)

Total entropy change,
∆S_total = ∆S_iron + ∆S_water
∆S_total = m_iron*c_iron*ln(Tf/T1) + m_water*c*ln(Tf/Ti)
∆S_total = 1*0.45*ln(306.15/1173.15) + 4*4.18*ln(306.15/283.15)
∆S_total = - 0.6045 + 1.3076
∆S_total = 0.7031 kJ/K