A0.600kg sample of water is initially ice at temperature$\mathbf{‐}\mathbf{20}\mathbf{°}\mathbf{C}$ .What is the sample’s entropy change if its temperature is increased to$\mathbf{40}\mathbf{°}\mathbf{C}$ ?

The mass of the sample of the water is $\mathrm{m}=0.600\mathrm{kg}$.

The initial temperature is ${\mathrm{T}}_{\mathrm{i}}=-20°\mathrm{C}=253\mathrm{K}$.

Increase in the temperature is $40°\mathrm{C}=313\mathrm{K}$.

First, we have to find the change in the entropy for initial and final temperatures of ice and from the initial temperature and the final temperature of the water. By adding these values, we get the total change in the entropy for the ice and water.

Formulae:

The energy received as heat when the temperature changes by $\mathrm{dT},\mathrm{dQ}=\mathrm{mcdT}$ (1)

The change in the entropy of the sample, $∆\mathrm{S}=\int \frac{\mathrm{dQ}}{\mathrm{T}}$ (2)

At melting point, the energy leaving the ice as heat, $\mathrm{Q}={\mathrm{mL}}_{\mathrm{F}}$ (3)

where, ${\mathrm{L}}_{\mathrm{F}}$is the heat of fusion of ice.

The total change in the entropy for the ice and the water, $∆S=∆S_1+∆S_2+∆S_3$ (4)

As ice warms, the energy it receives as heat when the temperature changes by dT.

Substituting equation (1) in equation (2), we can get the change in entropy of the sample as follows: (where, m is the mass of ice and ${\mathrm{c}}_{\mathrm{l}}$ is the specific heat of ice.)

$$\begin{gathered} ∆S_1={\mathrm{mc}}_{\mathrm{l}}{\int }_{{\mathrm{T}}_{\mathrm{l}}}^{{\mathrm{T}}_{\mathrm{f}}}\frac{\mathrm{dT}}{\mathrm{T}} \\ =\mathrm{mcln}\left(\frac{{\mathrm{T}}_{\mathrm{f}}}{{\mathrm{T}}_{\mathrm{i}}}\right) \\ =0.60\mathrm{kg}\times 2220\frac{\mathrm{J}}{\mathrm{kg}·\mathrm{K}}\times \mathrm{In}\left(\frac{273\mathrm{K}}{253\mathrm{K}}\right) \\ =732\mathrm{J}/\mathrm{K} \end{gathered}$$

We know that melting is an isothermal process. The entropy change is given by substituting equation (3) in equation (2) as follows: (where,${\mathrm{L}}_{\mathrm{F}}$is the heat of fusion of ice.)

$$\begin{gathered} ∆S_2=\frac{{\mathrm{mL}}_{\mathrm{F}}}{\mathrm{T}} \\ =\frac{0.60\mathrm{kg}\times 333\times {10}^3\mathrm{J}/\mathrm{kg}}{273\mathrm{K}} \\ =732\mathrm{J}/\mathrm{K} \end{gathered}$$

For the warming of the water from the melted ice, the change in the entropy is given by substituting equation (1) in equation (2) as follows: (where, $c_w$is the specific heat of the water.)

$$\begin{gathered} ∆S_3={\mathrm{mc}}_{\mathrm{w}}{\int }_{{\mathrm{T}}_{\mathrm{I}}}^{{\mathrm{T}}_{\mathrm{f}}}\frac{\mathrm{dT}}{\mathrm{T}} \\ ={\mathrm{mc}}_{\mathrm{w}}\mathrm{In}\left(\frac{{\mathrm{T}}_{\mathrm{f}}}{{\mathrm{T}}_{\mathrm{I}}}\right) \\ =0.6000\mathrm{kg}\times 4190\frac{\mathrm{J}}{\mathrm{kg}·\mathrm{K}}\times \mathrm{In}\left(\frac{313\mathrm{K}}{273\mathrm{K}}\right) \\ =344\mathrm{J}/\mathrm{K} \end{gathered}$$.

Therefore, the total change in the entropy for the ice and the water is given by equation (4) as follows:

$$\begin{gathered} ∆S=101\mathrm{J}/\mathrm{K}+732\mathrm{J}/\mathrm{K}+344\mathrm{J}/\mathrm{K} \\ =1.18\times {10}^3\mathrm{J}/\mathrm{K} \end{gathered}$$

From the above solution, we can see that the largest increase in the entropy comes from role="math" localid="1661745789809" $∆S_2$, which accounts for the melting process and the change in entropy is $1.18\times {10}^3\mathrm{J}/\mathrm{K}$