Answer
See explanation.
Work Step by Step
(a) The heat transfer between the water of 100°C and the 30°C water. The temperature is finite. It is a natural process since heat travels from hot place to cold place and does not require engines. This means the process is irreversible.
(b) The final temperature of bath water
$ m_1c(T_{f}-T_{i1}) =-m_2c(T_{f}-T_{i2}) $ Now we solve for $T_f$
$\frac{m_1}{m_2}=-\frac{(T_{f}-T_{i2}) }{ (T_{f}-T_{i1}) }$
$\frac{195kg}{5.0 kg}=-\frac{(T_{f}-100^oC) }{ (T_{f}-30^oC) }$
$39=-\frac{(T_{f}-100^oC) }{ (T_{f}-30^oC) }$
$39T_{f} - 1170 ^oC=-T_{f}+100^oC$
$39T_{f} + T_{f}=100^oC + 1170 ^oC$
$40T_{f} =1270 ^oC$
$T_{f} =\frac{3870 ^oC}{40}$
$T_{f} = 31.75 ^oC = 304.9 K$
(c) $\Delta S = m_1c ln(\frac{T_2}{T_1}) + m_2c ln(\frac{T_2}{T_1}) $
$\Delta S = (195 kg) (4190 J/kg.K) ln(\frac{304.9 K}{303.15 K}) + (5.0kg)(4190 J/kg.K) ln(\frac{304.9 K}{373.15K}) $
$\Delta S = 4703 J/K + (-4232 J/K) $
$\Delta S = 471 J/K$
