91Ó°ÊÓ

We can use the condition of equilibrium to find the final temperature of the system. Using the entropy change formula, we can find the entropy change for the whole process.

Formulae:

The heat released by the gas due to latent heat,

$\mathrm{Q}={\mathrm{mL}}_{\mathrm{F}}$ (1)

The heat absorbed by the gas,

$\mathrm{Q}=\mathrm{mcdT}$ (2)

The entropy of gas using the second law of thermodynamics,

$\mathrm{S}=\frac{\mathrm{Q}}{\mathrm{T}}$ (3)

The entropy of gas for a reversible process,

$\mathrm{S}=\mathrm{m}\mathrm{cIn}\left(\frac{{\mathrm{T}}_{\mathrm{final}}}{{\mathrm{T}}_{\mathrm{initial}}}\right)$ (4)

The heat required to change the state of m gram of ice using equation (1) is given as: (will use${\mathrm{L}}_{\mathrm{f}}=333×{10}^3\mathrm{J}/\mathrm{kg}$)

$$\begin{gathered} {\mathrm{Q}}_1=0.012\mathrm{kg}×333×{10}^3\mathrm{J}/\mathrm{kg} \\ =3996\mathrm{J} \end{gathered}$$

Heat transferred by the water using equation (2) is given as:

$$\begin{gathered} {\mathrm{Q}}_2=0.13\mathrm{kg}×4190\mathrm{J}/\mathrm{kg}.°\mathrm{C}×\left({\mathrm{T}}_{\mathrm{final}}-80°\mathrm{C}\right) \\ =\left(544.7\mathrm{J}/°\mathrm{C}\right){\mathrm{T}}_{\mathrm{final}}-43576\mathrm{J} \end{gathered}$$

Heat absorbed by the ice using equation (2) is given as:

$$\begin{gathered} {\mathrm{Q}}_3=0.012\mathrm{kg}×4190\mathrm{J}/\mathrm{kg}.°\mathrm{C}\left({\mathrm{T}}_{\mathrm{finial}}-0°\mathrm{C}\right) \\ =\left(50.28\mathrm{J}/°\mathrm{C}\right){\mathrm{T}}_{\mathrm{final}} \end{gathered}$$

Now use the condition

Substituting all values calculated in the above relation, we get that

$$\begin{gathered} \underset{}{∑\mathrm{Q}=0} \\ 3669\mathrm{J}+\left(50.28\mathrm{J}/°\mathrm{C}\right){\mathrm{T}}_{\mathrm{final}}+\left(544.7\mathrm{J}/°\mathrm{C}\right){\mathrm{T}}_{\mathrm{final}}-43576\mathrm{J}=0 \\ \left(594.44\mathrm{J}/°\mathrm{C}\right){\mathrm{T}}_{\mathrm{final}}=39580\mathrm{J} \\ {\mathrm{T}}_{\mathrm{final}}=66.58°\mathrm{C} \\ {\mathrm{T}}_{\mathrm{final}}=339.67\mathrm{K} \end{gathered}$$

Hence, the value of the equilibrium temperature is

The process of ice at $0°\mathrm{C}$turning to water at$0°\mathrm{C}$involves entropy change as given:

(The heat required to change the state ofa gram of ice is$\mathrm{Q}={\mathrm{mL}}_{\mathrm{F}}$, hence substituting the value of equation (1) in equation (3), we can get the entropy change)

Hence, the value of the entropy change when the ice melts is 14.6 J/K

The process of m = 0.012 kg of water warming from $0°\mathrm{C}\mathrm{to}66.5°\mathrm{C}$involves entropy change that is given by integrating the equation (3) as given:

$$\begin{gathered} \mathrm{S}={∫}_{{\mathrm{T}}_{\mathrm{i}}}^{{\mathrm{T}}_{\mathrm{f}}}\frac{\mathrm{dQ}}{\mathrm{T}} \\ ={∫}_{0°\mathrm{C}}^{66.5°\mathrm{C}}\frac{\mathrm{dQ}}{\mathrm{T}} \\ ={∫}_{273.15\mathrm{K}}^{33.67\mathrm{K}}\frac{\mathrm{mCdT}}{\mathrm{T}} \\ \mathrm{S}=\mathrm{mC}{∫}_{273.15\mathrm{K}}^{339.67\mathrm{K}}\frac{\mathrm{dT}}{\mathrm{T}} \\ =\mathrm{mCIn}\left(\frac{339.67\mathrm{K}}{273.15\mathrm{K}}\right) \\ =0.012\mathrm{kg}×4190\mathrm{J}/\mathrm{kg}.\mathrm{K}×0.218 \\ =11\mathrm{J}/\mathrm{K} \end{gathered}$$

Hence, the value of the entropy change is 11 J/K

As in process (c), the cooling of original water $\mathrm{m}=0.130\mathrm{kg}\mathrm{from}80°\mathrm{C}\mathrm{to}66.5°\mathrm{C}$involves entropy change using equation (iv) as given:

$$\begin{gathered} \mathrm{S}=0.130\mathrm{kg}×4190\mathrm{J}/\mathrm{kg}.\mathrm{K}×\mathrm{In}\left(\frac{339.67\mathrm{K}}{353.15\mathrm{K}}\right) \\ ==21.2\mathrm{J}/\mathrm{K} \end{gathered}$$

Hence, the value of the entropy change is -21.2 J/K

Net entropy is found by summing all the entropies calculated above as given:

$$\begin{gathered} ∆{\mathrm{S}}_{\mathrm{NET}}=\left(14.6+11-21.2\right)\mathrm{J}/\mathrm{K} \\ =4.4\mathrm{J}/\mathrm{K} \end{gathered}$$

Hence, the value of the net entropy change is 4.4 J/K