Question

Question: Calculate the entropy change that results from mixing $\mathbf{54}\mathbf{.}\mathbf{0}\mathit{g}$ water at$\mathbf{273}\mathit{K}$ with $\mathbf{27}\mathbf{.}\mathbf{0}\mathit{g}$water at$\mathbf{373}\mathit{K}$ in a vessel whose walls are perfectly insulated from the surroundings. Consider the specific heat of water to be constant over the temperature range from 273 to $\mathbf{373}\mathit{K}$and to have the value $\mathbf{4}\mathbf{.18}\mathbf{}\mathbf{\text{J}}\mathbf{}{\mathbf{\text{K}}}^{\mathbf{-}\mathbf{1}}\mathbf{}{\mathbf{\text{g}}}^{\mathbf{-}\mathbf{1}}$.

Short answer

The total entropy change $\Delta S=3.78\text{J}{\text{K}}^{-1}$.

Step-by-step solution

Given data

The mixing water of different masses and temperatures in an isolated vessel:

$m_1\left({\text{H}}_2\text{O}\right)=54.0\text{g\hspace{0.33em}}{T}_1=273\text{K}$.

$m_2\left({\text{H}}_2\text{O}\right)=27.0\text{g\hspace{0.33em}}{T}_2=373\text{K}$.

Capacity $c_p=4.18\text{J}{\text{K}}^{-1}{\text{g}}^{-1}$.

Concept of entropy

Entropy Change is the phenomenon that is the measure of change of disorder or randomness in a thermodynamic system. It is related to the conversion of heat or enthalpy in work. A thermodynamic system that has more randomness means it has high entropy.

The formula to calculate the entropy change

The total entropy change will be a sum of the separate entropy changes, 1 of each mixture:

$\Delta S=\Delta S_1+\Delta S_2$ .......... (1)

The formula for the entropy change is:

$\Delta S=m{C}_s\ln \frac{T_{\text{final}}}{T_{\text{initial}}}$ ............ (2)

The final temperature will be somewhere between 273 and 373K . This is the only information that we need to calculate in order to determine the entropy change.

Calculate the final temperature

The final temperature will be obtained from the formula for the heat:

$q=C_{s}m\Delta T$

This would need to be calculated for each process:

$\begin{array}{l}{q}_1=m_1{C}_s\left(T_{\text{final}}-T_1\right)\\ q_2=m_2{C}_s\left(T_{\text{final}}-T_2\right)\end{array}$

$q_1=54.0\times 4.18\left(T_{\text{final}}-273\right)$ ......... (3)

$q_2=27.0\times 4.18\left(T_{\text{final}}-373\right)$ .......... (4)

The system is isolated, therefore the overall heat transferred will be .

This would written as: $q=q_1+q_2=0$.

Combine the equation (3) and (4):

$\begin{array}{c}54.0\times 4.18\left(T_{\text{final}}-273\right)+27.0\times 4.18\left(T_{\text{final}}-373\right)=0\\ 225.72T_{\text{final}}-61621.56+112.86T_{\text{final}}-42096.78=0\\ T_{\text{final}}=103718.34÷338.58\\ =306.33\text{K}\end{array}$

Calculate the value of ΔS1

Calculate $\Delta S_1$ as follows:

$\begin{array}{l}{m}_1\left({\text{H}}_2\text{O}\right)=54.0\text{g\hspace{0.33em}}\\ T_i=273\text{K}\\ T_f=306.33K\\ c_p=4.18\text{J}{\text{K}}^{-1}{\text{g}}^{-1}\end{array}$

Substitute these values in equation (2) to calculate the entropy change:

$\begin{array}{c}\Delta S_1=54.00\times 4.18\times \ln \frac{306.33}{273}\\ =26.00\text{J}{\text{K}}^{-1}\end{array}$

Calculate the value of  ΔS2

Calculate $\Delta S_2$ as follows:

$\begin{array}{l}{m}_1\left({\text{H}}_2\text{O}\right)=27.0\text{g\hspace{0.33em}}\\ T_i=373\text{K}\\ T_f=306.33K\\ c_p=4.18\text{J}{\text{K}}^{-1}{\text{g}}^{-1}\end{array}$

Substitute these values in equation (2) to calculate the entropy change:

$\begin{array}{c}\Delta S_2=27.00\times 4.18\times \ln \frac{306.33}{373}\\ =-22.22\text{J}{\text{K}}^{-1}\end{array}$

Calculate the overall entropy change

$$\begin{gathered} \Delta S_1=26.00\text{J}{\text{K}}^{-1} \\ \Delta S_2=-22.22\text{J}{\text{K}}^{-1} \end{gathered}$$

Substitute these values in equation (1)

$\begin{array}{c}\Delta S=26.00-22.22\\ =3.78\text{J}{\text{K}}^{-1}\end{array}$

$\begin{array}{c}\Delta S=26.00-22.22\\ =3.78\text{J}{\text{K}}^{-1}\end{array}$