91Ó°ÊÓ

Mass of ice is$m_{ice}=8gm\left(\frac{{10}^{-3}\mathrm{kg}}{1\mathrm{gm}}\right)=0.008\mathrm{kg}$

Initial Temperature of ice${\mathrm{T}}_{\mathrm{initial}}^{\mathrm{ice}}=-10°\mathrm{C}=263.15\mathrm{kg}$

Specific heat of ice${\mathrm{c}}_{\mathrm{ice}}=2220\mathrm{J}/\lg .\mathrm{K}=2220\mathrm{J}/\mathrm{kg}.°\mathrm{C}$

Mass of water${\mathrm{m}}_{\mathrm{water}}=100\mathrm{gm}\left(\frac{{10}^{-3}\mathrm{kg}}{1\mathrm{gm}}\right)=0.1\mathrm{kg}$

Initial Temperature of water${\mathrm{T}}_{\mathrm{initial}}^{\mathrm{water}}=20°\mathrm{C}=293.15\mathrm{K}$

We use the concept of entropy change of the two blocks coming to thermal equilibrium to calculate entropy. We use the formula of specific heat of the water to calculate the final temperature of the water.

Formulae:

The entropy change of the gas,

$∆{\mathrm{S}}_{\mathrm{temp}\mathrm{change}}=\mathrm{mc}\mathrm{In}\left(\frac{{\mathrm{T}}_{\mathrm{final}}}{{\mathrm{T}}_{\mathrm{initial}}}\right)$ (1)

$∆\mathrm{S}=\frac{{\mathrm{L}}_{\mathrm{F}}\mathrm{m}}{\mathrm{T}}$

The heat transferred by the body,

$\mathrm{Q}=\mathrm{cm}∆\mathrm{T}$ (2)

$0.64\mathrm{J}/°\mathrm{C}$For an equilibrium state, the heat lost by $100{\mathrm{cm}}^3\left({\mathrm{m}}_{\mathrm{water}}\right)$of water is absorbed by ice$\left({\mathrm{m}}_{\mathrm{ice}}\right)$which melts and reaches temperature${\mathrm{T}}_{\mathrm{f}}>0$

For${\mathrm{m}}_{\mathrm{water}}=0.1\mathrm{kg}$of water, the specific heat is given by

$$\begin{gathered} {\mathrm{c}}_{\mathrm{water}}=4.190\mathrm{J}/\mathrm{g}.°\mathrm{C} \\ =4190\mathrm{J}/\mathrm{kg}.°\mathrm{C} \end{gathered}$$

When the equilibrium is reached, then$\underset{}{∑}Q=0$

During the equilibrium process, warm waterat$20°C$cools to a final temperature${\mathrm{T}}_{\mathrm{f}}$, ice at temperature$-10°\mathrm{C}$warms to$0°C$, and this ice first melts, and then melted ice (water) warms.

${\mathrm{Q}}_{\mathrm{warm}\mathrm{water}\mathrm{cools}}+{\mathrm{Q}}_{\mathrm{ice}\mathrm{warms}\mathrm{to}0°\mathrm{C}}+{\mathrm{Q}}_{\mathrm{ice}\mathrm{melts}}+{\mathrm{Q}}_{\mathrm{melted}\mathrm{ice}\mathrm{warms}=0}$

The specific heat of water is given by the formula of equation (2) as given:

$c_{water}{m}_{water}(T_f-20°C)+c_{ice}{m}_{ice}(0°C-(-10°C\left)\right)+L_F{m}_{ice}+c_{water}{m}_{ice}(T_f-0°C)=0$

${\mathrm{L}}_{\mathrm{F}}$is heat fusion of water, and its value is given by$L_F=333×{10}^3\frac{J}{kg}$

Substituting all values in the above expression, we get that the final temperature can be obtained as,

$$\begin{gathered} \left[\left(4190\mathrm{J}/\mathrm{kg}.°\mathrm{C}×0.1\mathrm{kg}\right){\mathrm{T}}_{\mathrm{f}}-\left(4190\mathrm{J}/\mathrm{kg}.°\mathrm{C}×0.1\mathrm{kg}×20°\mathrm{C}\right)\right]+ \\ \left[2220\mathrm{J}/\mathrm{kg}.°\mathrm{C}×0.008\mathrm{kg}×10°\mathrm{C}\right]+\left(333×{10}^3\mathrm{J}/\mathrm{kg}×0.008\mathrm{kg}\right) \\ +\left(4190\mathrm{J}/\mathrm{kg}.°\mathrm{C}×0.008\mathrm{kg}×{\mathrm{T}}_{\mathrm{f}}\right)=0 \end{gathered}$$

$$\begin{gathered} \left(419\mathrm{J}/°\mathrm{C}\right){\mathrm{T}}_{\mathrm{f}}-8380\mathrm{J}+177.6\mathrm{J}+2841.6\mathrm{J}+\left(33.52\mathrm{J}/°\mathrm{C}\right){\mathrm{T}}_{\mathrm{f}}=0 \\ \left(452.52\mathrm{J}/°\mathrm{C}\right){\mathrm{T}}_{\mathrm{f}}=5538.4\mathrm{J} \\ {\mathrm{T}}_{\mathrm{f}}=12.24°\mathrm{C} \\ {\mathrm{T}}_{\mathrm{f}}=285.39\mathrm{K} \end{gathered}$$

Entropy change of two blocks coming to equilibrium is given by using equation (i) can be given as:

$$\begin{gathered} ∆{\mathrm{S}}_{\mathrm{temp}\mathrm{change}}=\mathrm{mcIn}\left(\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}\right) \\ ∆{\mathrm{S}}_{\mathrm{melt}}=\frac{{\mathrm{L}}_{\mathrm{F}}\mathrm{m}}{{\mathrm{T}}_0} \end{gathered}$$

For the phase change experienced by the ice$(with{T}_0=273.15 K)$the total entropy change is given as:

$$\begin{gathered} ∆{\mathrm{S}}_{\mathrm{system}}={\mathrm{m}}_{\mathrm{water}}{\mathrm{c}}_{\mathrm{water}}\mathrm{In}\left(\frac{285.39\mathrm{K}}{293.15\mathrm{K}}\right)+{\mathrm{m}}_{\mathrm{ice}}{\mathrm{c}}_{\mathrm{ice}}\mathrm{In}\left(\frac{273.15\mathrm{K}}{263.15\mathrm{K}}\right) \\ +{\mathrm{m}}_{\mathrm{ice}}{\mathrm{c}}_{\mathrm{water}}\mathrm{In}\left(\frac{285.39\mathrm{K}}{273.15\mathrm{K}}\right)+\left(\frac{{\mathrm{L}}_{\mathrm{F}}{\mathrm{m}}_{\mathrm{ice}}}{273.15\mathrm{K}}\right) \\ =\left(0.1\mathrm{kg}\right)\left(4190\mathrm{J}/\mathrm{kg}.°\mathrm{C}\right)\mathrm{In}\left(\frac{285.39\mathrm{K}}{293.15\mathrm{K}}\right)+ \\ \left(0.008\mathrm{kg}\right)\left(2220\mathrm{J}/\mathrm{kg}.°\mathrm{C}\right)\mathrm{In}\left(\frac{285.39\mathrm{K}}{273.15\mathrm{K}}\right)+\left(\frac{2664\mathrm{J}}{273.15\mathrm{K}}\right) \end{gathered}$$

$$\begin{gathered} ∆S_{system}=\left(-11.24+0.66+1.47+9.75\right)J/°C \\ =0.64J/°C \end{gathered}$$

Hence, the value of the entropy change of the system is$0.64\mathrm{J}/°\mathrm{C}$