91Ó°ÊÓ

Amount of hot water $=V_1=50L=50000g$

Temperature of hot water $=T_1=55°C=328K$

Amount of cold water $=V_2=25L=25000g$

Temperature of cold water $=T_2=10°C=283K$

Since the hot and cold waters are mixed together, the final temperature of the water can be given as:

$T_f=\frac{\left(T_1×V_1\right)+\left(T_2×V_2\right)}{V_1+V_2}$

By substituting the values in the above equation, we get,

$$\begin{gathered} T_f=\frac{(328×50)+(283×25)}{50+25} \\ {T}_f=313K \end{gathered}$$

The change in entropy is given as:

role="math" localid="1647236968068" $\mathrm{Δ}S=C_V{∫}_{T_i}^{T_f}\frac{1}{T}dT..........\left(1\right)$

Where,

$C_V$ is the heat capacity at constant volume and is given as:
$C_V=mc$

Where,

$m$= mass

$c$= specific heat

Hence, equation (1) can be written in a simplified way as:

$\mathrm{Δ}S=mc\left[\ln \frac{T_f}{T_i}\right]..........\left(2\right)$

Now,

for hot water:

$$\begin{gathered} T_i=328K \\ m=50000g \end{gathered}$$

By substituting the values in equation (2), we get,

role="math" localid="1647238076045" $$\begin{gathered} \mathrm{Δ}{S}_{hot}=50000×4.18×\left[\ln \frac{313}{328}\right] \\ \mathrm{Δ}{S}_{hot}=-9783.38J{K}^{-1} \end{gathered}$$

for cold water:

$$\begin{gathered} T_i=283K \\ m=25000g \end{gathered}$$

By substituting the values in equation (2), we get,

$$\begin{gathered} \mathrm{Δ}{S}_{cold}=25000×4.18×\left[\ln \frac{313}{283}\right] \\ \mathrm{Δ}{S}_{cold}=10529.03J{K}^{-1} \end{gathered}$$

Thus, the net entropy change can be given as:

$$\begin{gathered} \mathrm{Δ}S=\mathrm{Δ}{S}_{hot}+\mathrm{Δ}{S}_{cold} \\ \mathrm{Δ}S=-9783.38+10529.03 \\ \mathrm{Δ}S=745.65J{K}^{-1} \end{gathered}$$

Hence, the required change in entropy can be calculated as$745.65J{K}^{-1}$.