Respuesta :
Solution :
a). Applying the energy balance,
[tex]$\Delta E_{sys}=E_{in}-E_{out}$[/tex]
[tex]$0=\Delta U$[/tex]
[tex]$0=(\Delta U)_{iron} + (\Delta U)_{water}$[/tex]
[tex]$0=[mc(T_f-T_i)_{iron}] + [mc(T_f-T_i)_{water}]$[/tex]
[tex]$0 = 27 \times 0.45 \times (T_f - 375) + 130 \times 4.18 \times (T_f-26)$[/tex]
[tex]$t_f=33.63^\circ C$[/tex]
b). The entropy change of iron.
[tex]$\Delta s_{iron} = mc \ln\left(\frac{T_f}{T_i} \right)$[/tex]
[tex]$ = 27 \times 0.45\ \ln\left(\frac{33.63 + 273}{375 + 273} \right)$[/tex]
= -9.09 kJ-K
Entropy change of water :
[tex]$\Delta s_{water} = mc \ \ln\left(\frac{T_f}{T_i} \right)$[/tex]
[tex]$ = 130 \times 4.18\ \ln\left(\frac{33.63 + 273}{26 + 273} \right)$[/tex]
= 10.76 kJ-K
So, the total entropy change during the process is :
[tex]$\Delta s_{tot} = \Delta s_{iron} + \Delta s_{water} $[/tex]
= -9.09 + 10.76
= 1.67 kJ-K
c). Exergy of the combined system at initial state,
[tex]$X=(U-U_{0}) - T_0(S-S_0)+P_0(V-V_0)$[/tex]
[tex]$X=mc (T-T_0) - T_0 \ mc \ \ln \left(\frac{T}{T_0} \right)+0$[/tex]
[tex]$X=mc\left((T-T_0)-T_0 \ ln \left(\frac{T}{T_0} \right)\right)$[/tex]
[tex]$X_{iron, i} = 27 \times 0.45\left(((375+273)-(12+273))-(12+273) \ln \frac{375+273}{12+273}\right)$[/tex]
[tex]$X_{iron, i} =63.94 \ kJ$[/tex]
[tex]$X_{water, i} = 130 \times 4.18\left(((26+273)-(12+273))-(12+273) \ln \frac{26+273}{12+273}\right)$[/tex]
[tex]$X_{water, i} =-13.22 \ kJ$[/tex]
Therefore, energy of the combined system at the initial state is
[tex]$X_{initial}=X_{iron,i} +X_{water, i}$[/tex]
= 63.94 -13.22
= 50.72 kJ
Similarly, Exergy of the combined system at initial state,
[tex]$X=(U_f-U_{0}) - T_0(S_f-S_0)+P_0(V_f-V_0)$[/tex]
[tex]$X=mc\left((T_f-T_0)-T_0 \ ln \left(\frac{T_f}{T_0} \right)\right)$[/tex]
[tex]$X_{iron, f} = 27 \times 0.45\left(((33.63+273)-(12+273))-(12+273) \ln \frac{33.63+273}{12+273}\right)$[/tex]
[tex]$X_{iron, f} = 216.39 \ kJ$[/tex]
[tex]$X_{water, f} = 130 \times 4.18\left(((33.63+273)-(12+273))-(12+273) \ln \frac{33.63+273}{12+273}\right)$[/tex]
[tex]$X_{water, f} =-9677.95\ kJ$[/tex]
Thus, energy or the combined system at the final state is :
[tex]$X_{final}=X_{iron,f} +X_{water, f$[/tex]
= 216.39 - 9677.95
= -9461.56 kJ
d). The wasted work
[tex]$X_{in} - X_{out}-X_{destroyed} = \Delta X_{sys}$[/tex]
[tex]$0-X_{destroyed} = $[/tex]
[tex]$X_{destroyed} = X_{initial} - X_{final}$[/tex]
= 50.72 + 9461.56
= 9512.22 kJ
