Question

An 10 gice cube at$\mathbf{-}\mathbf{10}\mathbf{°}\mathit{C}$is placed in a lake whose temperature is$\mathbf{15}\mathbf{°}\mathit{C}$. Calculate the change in entropy of the cube–lake system as the ice cube comes to thermal equilibrium with the lake. The specific heat of ice is 2220J/kg.K. (Hint: Will the ice cube affect the lake temperature?)

Short answer

The change in entropy of the cube-lake system as the ice cube comes to thermal equilibrium with the lake is 0.76J/K.

Step-by-step solution

The given data

a) Mass of the ice cube,$m=10gor0.010kg$

b) Temperature of the ice-cube,$T_l=-10°Cor274K$

c) Temperature of the lake,$T_W=15°C=288k$

d) Specific heat of ice,$c_l=2220J/Kg.k$

e) Specific heat of water,$c_w=-4190J/kg.k$

Understanding the concept of entropy change

Entropy change is a phenomenon that quantifies how disorder or randomness has changed in a thermodynamic system. The total change in entropy of the ice is the sum of entropy change while reaching the freezing point during the phase change and after it. We have to find the total change in entropy of the ice using the corresponding formulae. Then, we have to find the change in entropy of the lake water during the above processes. Adding both will give the total change in entropy of thecube-lake system.

Formulae:

The entropy change of a gas,$∆S=mc\ln \frac{T_f}{T_i}or∆S=\frac{Q}{T}$ …(i)

The heat released by the body due to latent heat,$Q=mL$ …(ii)

The heat transferred by the body, $Q=-m{C}_l\left(T_f-T_i\right)$ …(iii)

Calculation of the entropy change of cube-lake system

Initially, the temperature of the ice cube increases from$-10°Cto0°$

The change in entropy during this can be calculated using equation as follows:

$$\begin{gathered} ∆S=0.010kg\left(2220J/kg.k\right)\ln \left(\frac{273K}{263K}\right) \\ =0.828J/K \end{gathered}$$

During the phase change, the change in entropy of ice using equation (i) can be given as:

($L_F$is the heat of fusion for ice and$L_F=333\times {10}^{3}J/kg$)

$$\begin{gathered} ∆S=\frac{0.010kg\times 333\times {10}^{3}J/kg}{273K} \\ =12.1978J/K \\ \approx 12.2J/K \end{gathered}$$

Due to phase change, the ice melts and converts to water. After that, it starts warming. The change in entropy during this process is

(Here$c_w$is the specific heat of water.)

$$\begin{gathered} ∆S=0.010kg\left(4190J/kg.K\right)\ln \left(\frac{288K}{273K}\right) \\ =2.2412J/K \\ \approx 2.24J/K \end{gathered}$$

Hence, total change entropy of ice is

role="math" localid="1661326138002" $$\begin{gathered} ∆S=0.828J/K+12.2J/K+2.24J/K \\ =15.27J/K................................\left(a\right) \end{gathered}$$

Since the melting of the ice cube does not affect the lake temperature significantly, the change in entropy is given using equation (ii).

The energy lost by the lake water during the increase in temperature of ice from$-10°Cto0°C$ is given using equation (iii) as:

$$\begin{gathered} Q=-0.010kg\times 2220J/kg.K\times 10K \\ =-222J \end{gathered}$$

The energy lost by the lake water during the phase change of the ice is given using equation (ii) as: (negative sign indicates the lost energy)

$$\begin{gathered} Q=-0.010kg\times 333\times {10}^{3}J/kg \\ =-3.33\times {10}^{3}J \end{gathered}$$

The energy lost by the lake water after phase change of the ice using equation (iii) is given as:

$$\begin{gathered} Q=-0.010kg\times 4190J/kg.K\times 15K \\ =-628.5J \\ \approx -629J \end{gathered}$$

Thus, the total energy lost by the lake water is given by:

$$\begin{gathered} Q=-222J-3.33\times {10}^{3}J-629J \\ =-4181J \end{gathered}$$

Hence, the change in entropy of the lake water using equation (i) is given as:

$$\begin{gathered} ∆S=-\frac{4181J}{288K} \\ =-14.51J/K................\left(b\right) \end{gathered}$$

Total change in entropy of the ice-lake system using values from equations (a) and (b) is given as:

$$\begin{gathered} ∆S=15.27J/K-14.15J/K \\ \approx 0.76J/K \end{gathered}$$

Hence, the entropy change of the system is$0.76J/K$