91影视

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

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

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

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

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

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,$鈭�S=mc\ln \frac{T_f}{T_i}or鈭�S=\frac{Q}{T}$ 鈥�(颈)

The heat released by the body due to latent heat,$Q=mL$ 鈥�(颈颈)

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

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

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

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

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$L_F=333脳{10}^{3}J/kg$)

$$\begin{gathered} 鈭�S=\frac{0.010kg脳333脳{10}^{3}J/kg}{273K} \\ =12.1978J/K \\ 鈮�12.2J/K \end{gathered}$$

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.)

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

Hence, total change entropy of ice is

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

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$-10掳Cto0掳C$ is given using equation (iii) as:

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

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)

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

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

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

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

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

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

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

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

$$\begin{gathered} 鈭�S=15.27J/K-14.15J/K \\ 鈮�0.76J/K \end{gathered}$$

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