Question

An insulated Thermos contains 130gof water at 80.0$°$C. You put in an12.0 g ice cube at$\mathbf{0}\mathbf{°}\mathit{C}$to form a system of ice + original water. (a) What is the equilibrium temperature of the system? What are the entropy changes of the water that was originally the ice cube (b)as it melts and (c)as it warms to the equilibrium temperature? (d)What is the entropy change of the original water as it cools to the equilibrium temperature? (e)What is the net entropy change of the ice + original water system as it reaches the equilibrium temperature?

Short answer

a) Equilibrium temperature of the system is 339.67 K.

b) Entropy changes of the water that was originally the ice cube as it melts is 14.6 J/K.

c) Entropy changes of the water that were originally the ice cube as it warms to the equilibrium temperature is 11 J/K

d) Entropy change of the original water as it cools to the equilibrium temperature is -21.2 J/K.

e) Net entropy change of the ice+ original water system as it reaches the equilibrium is 4.0 J/K

Step-by-step solution

The given data

$$\begin{gathered} \mathrm{Mass}\mathrm{of}\mathrm{water}{\mathrm{m}}_{\mathrm{w}}=130\mathrm{gm}\left(\frac{{10}^{-3}\mathrm{kg}}{1\mathrm{gm}}\right)=0.130\mathrm{kg} \\ \mathrm{Temperature}\mathrm{of}\mathrm{water}{\mathrm{T}}_{\mathrm{w}}=80°\mathrm{C}=353\mathrm{K} \\ \mathrm{Mass}\mathrm{of}\mathrm{ice}\mathrm{cube}{\mathrm{m}}_{\mathrm{ice}}=12\mathrm{gm}\left(\frac{{10}^{-3}\mathrm{kg}}{1\mathrm{gm}}\right)=0.012\mathrm{kg} \\ \mathrm{The}\mathrm{temperature}\mathrm{of}\mathrm{the}\mathrm{ice}\mathrm{cube}{\mathrm{T}}_{\mathrm{ice}}=0°\mathrm{C}=273\mathrm{K} \end{gathered}$$

Understanding the concept of entropy change

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)

(a) Calculation of the equilibrium temperature of the system

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

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

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

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

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

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

Now use the condition

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

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

Hence, the value of the equilibrium temperature is

(b) Calculation of entropy changes of the water that was originally the ice cube as it melts

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)

$$\begin{gathered} \mathrm{S}=\frac{{\mathrm{mL}}_{\mathrm{f}}}{\mathrm{T}} \\ =\frac{0.012\mathrm{kg}\times 33\times {10}^3\mathrm{J}/\mathrm{kg}}{273.15\mathrm{K}} \\ =14.6\mathrm{J}/\mathrm{K} \end{gathered}$$

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

(c) Calculation of the entropy changes of the water that was originally the ice cube as it warms to the equilibrium temperature

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:

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

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

(d) Calculation of the entropy change of the original water as it cools to the equilibrium temperature

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:

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

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

(e) Calculation of the net entropy change of the ice + original water system

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