Question 5.1: A 40 kg steel casting (CP = 0.5 kJ⋅kg^−1.K^−1) at a temperat...

While the overall process described here is irreversible, one can imagine the cooling of the casting from its initial to final temperature by reversible heat transfer and the corresponding heating of the oil by reversible heat transfer. In each case, the entropy change is given by Eq. (5.1). If the oil and the casting are both incompressible,

d S^t=d Q_{ rev } / T       (5.1)

then in each case d Q_{ rev }=d U^t=C_p^t dT . For an incompressible substance, all processes are constant volume processes, so C_{p} and C_{v} are equal and no reversible work is possible. Integrating Eq. (5.1) from an initial temperature T_{1}  to a final temperature T{2},   for  constant  C_{P}^{t}  , then gives the change in entropy as:

\Delta S^t=\int_{T_1}^{T_2} \frac{d Q_{ rev }}{T}=\int_{T_1}^{T_2} \frac{C_p^t d T}{T}=C_p^t \ln \frac{T_2}{T_1}

The final temperature t of the oil and the steel casting is found by an energy balance. Because the change in energy of the oil and steel together must be zero,

(40)(0.5)(t − 450) + (150)(2.5)(t − 25) = 0

Solution yields t = 46.52°C.

(a) Change in entropy of the casting:

\Delta S^t=m \int \frac{C_P d T}{T}=m C_P \ln \frac{T_2}{T_1}

=(40)(0.5) \ln \frac{273.15+46.52}{273.15+450}=-16.33  kJ \cdot K ^{-1}

(b) Change in entropy of the oil:

\Delta S^t=(150)(2.5) \ln \frac{273.15+46.52}{273.15+25}=26.13  kJ \cdot K ^{-1}

(c) Total entropy change:

Note that although the total entropy change is positive, the entropy of the casting has decreased.