A 600 glump of copper at $\mathbf{80}\mathbf{.}\mathbf{0}\mathbf{°}\mathbf{C}$ is placed in$\mathbf{70}\mathbf{.}\mathbf{0}\mathit{g}$of water at$\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{°}\mathit{C}$in an insulated container. (See Table 18-3 for specific heats.) (a) What is the equilibrium temperature of the copper water system? What entropy changes do (b) the copper, (c) the water, and (d) the copper–water system undergo in reaching the equilibrium temperature?

The mass of copper lump,${\mathrm{m}}_{\mathrm{c}}=600\mathrm{g}$

The initial temperature of copper mass,${\mathrm{T}}_{\mathrm{ice}}=80.0°\mathrm{C}=353\mathrm{K}$

The mass of water, $m_w=70.0g$

The initial temperature of water, ${\mathrm{T}}_{\mathrm{iwater}}=10.0°\mathrm{C}=283\mathrm{K}$

The specific heat of copper,$c_c=0.092cal/g·K$

The specific heat of water,$c_w=1.00cal/g·K$

The hot copper lump, when dropped in cold water, will lose its excess heat. The copper-water system is isolated by an insulating container. Hence the heat lost by the copper lump will be absorbed by the water. This will raise the temperature of the water. Thus, the copper-water system will reach an equilibrium temperature. As the heat is transferred from one object to another, there will be a change in entropy.

Formulae:

The energy lost or absorbed as heat by the system, $\mathrm{Q}=\mathrm{mc}\left({\mathrm{T}}_{\mathrm{f}}-{\mathrm{T}}_{\mathrm{i}}\right)$ (1)

The entropy change of the gas at constant volume, $∆\mathrm{S}=\mathrm{mcln}\left(\frac{{\mathrm{T}}_{\mathrm{f}}}{{\mathrm{T}}_{\mathrm{i}}}\right)$ (2)

The heat lost by the copper lump using equation (1) is calculated as:

$Q_c=m_c{c}_c\left(T_f-T_{ice}\right)..................................\left(3\right)$

The heat absorbed by the water using equation (1) is calculated as:

role="math" localid="1661578703302" $Q_w=m_w{c}_w\left(T_f-T_{iwater}\right)..................................\left(4\right)$

As the total energy of the system cannot change, the sum of the heat transfers will be zero. Thus using equation (3) and (4), we get the equilibrium temperature as:

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{c}}+{\mathrm{Q}}_{\mathrm{w}}=0 \\ {\mathrm{m}}_{\mathrm{c}}{\mathrm{c}}_{\mathrm{c}}\left({\mathrm{T}}_{\mathrm{f}}-{\mathrm{T}}_{\mathrm{ice}}\right)+{\mathrm{m}}_{\mathrm{w}}{\mathrm{c}}_{\mathrm{w}}\left({\mathrm{T}}_{\mathrm{w}}-{\mathrm{T}}_{\mathrm{iwater}}\right) \\ {\mathrm{T}}_{\mathrm{f}}=\frac{{\mathrm{m}}_{\mathrm{c}}{\mathrm{c}}_{\mathrm{c}}{\mathrm{T}}_{\mathrm{ice}}+{\mathrm{m}}_{\mathrm{w}}{\mathrm{c}}_{\mathrm{w}}{\mathrm{T}}_{\mathrm{iwater}}}{{\mathrm{m}}_{\mathrm{c}}{\mathrm{c}}_{\mathrm{c}}+{\mathrm{m}}_{\mathrm{w}}{\mathrm{c}}_{\mathrm{w}}} \\ {\mathrm{T}}_{\mathrm{f}}=\frac{\left(600\mathrm{g}\times 0.092\mathrm{cal}/\mathrm{g}·\mathrm{K}\times 353\mathrm{K}\right)+\left(70.0\mathrm{g}\times 1.00\mathrm{cal}/\mathrm{g}·\mathrm{K}\times 283\mathrm{K}\right)}{\left(600\mathrm{g}\times 0.092\mathrm{cal}/\mathrm{g}·\mathrm{K}\right)\left(70.0\mathrm{g}\times 1.00\mathrm{cal}/\mathrm{g}·\mathrm{K}\right)} \\ {\mathrm{T}}_{\mathrm{f}}=\frac{19.49\times {10}^3\mathrm{cal}+19.81\times {10}^3\mathrm{cal}}{55.2\mathrm{cal}/\mathrm{K}+70.0\mathrm{cal}/\mathrm{k}} \\ {\mathrm{T}}_{\mathrm{f}}=\frac{39.3\times {10}^3\mathrm{cal}}{125.2\mathrm{cal}/\mathrm{K}} \\ {\mathrm{T}}_{\mathrm{f}}=313.9\mathrm{K} \\ {\mathrm{T}}_{\mathrm{f}}=40.9°\mathrm{C} \end{gathered}$$

Hence, the value of the equilibrium temperature of the system is$40.9°C$

Entropy of copper using equation (2) is calculated as given:

$$\begin{gathered} ∆S=600g\times 0.092cal/g·K\times \ln \left(\frac{313.9K}{353K}\right) \\ =600g\times 0.092cal/g·K\times \ln \left(0.89\right) \\ =-6.62cal/K \\ =-6.62cal\times \frac{4.1J}{1cal} \\ =-27.1J/K \end{gathered}$$

Hence, the entropy change of the copper is$-27.1J/K$

Entropy of water using equation (2) is calculated as given:

$$\begin{gathered} ∆S=70.0g\times 1.00cal/g·K\times \ln \left(\frac{313.9K}{283K}\right) \\ =70.0g\times 1.00cal/g·K\times \ln \left(1.1\right) \\ =7.82cal/K \\ =7.82cal\times \frac{4.1J}{1cal} \\ =30.3J/K \end{gathered}$$

Hence, the value of the entropy change of water is 30.3 J/K

The entropy of the copper-water system is the difference between the entropies of the copper and water individually. Hence, it is given as:

$$\begin{gathered} \mathrm{S}\left(\mathrm{system}\right)=30.3\mathrm{J}/\mathrm{K}-27.1\mathrm{J}/\mathrm{K} \\ =3.2\mathrm{J}/\mathrm{K} \end{gathered}$$

Hence, the entropy change of the system is 3.2 J/K