91Ó°ÊÓ

a) Mass of the water,$m_w=50.0\mathrm{g}\mathrm{or}50×{10}^{-3}\mathrm{kg}$

b) Mass of the aluminum,$m_{al}=200gor200×{10}^{-3}\mathrm{kg}$

c) Temperature of aluminum,$T_{al}=100°\mathrm{C}\mathrm{or}373\mathrm{K}$

d) Temperature of water,$T_w=20.0°\mathrm{C}\mathrm{or}293\mathrm{K}$

e) Specific heat of aluminum,$c_{al}=900\mathrm{J}/\mathrm{kg}.\mathrm{K}$

f) Specific heat of water,$c_{al}=4190\mathrm{J}/\mathrm{kg}.\mathrm{K}$

Entropy change is a phenomenon that quantifies how disorder or randomness has changed in a thermodynamic system. We can write the condition of equilibrium for heat energy absorbed and transferred by the two blocks, and by writing it in terms of specific heat and temperature; we can find the value ofequilibrium temperature of the two-block system. Then, by using the formula for change in entropy, we can find the change in entropy of the aluminum, water, and aluminum-water system.

Formulae:

The heat transferred by the body,$Q=mc∆T$ …(¾±)

The entropy change of the gas, $∆S=mcln\left(\frac{T_f}{T_i}\right)$ …(¾±¾±)

At equilibrium, the total heat of the system is zero. So,$Q_w+Q_{al}=0$

Using equation (i) and the given values, we can get that

role="math" localid="1661323263437" $m_w{c}_w∆T_w+m_{al}{c}_{al}∆T_{al}=0$

Let the equilibrium temperature be $T_f$Then, the above equation becomes

$$\begin{gathered} m_w{c}_w(T_f-T_w)+m_{al}{c}_{al}(T_f-T_{al}=0 \\ 50×{10}^{-3}\mathrm{kg}\left(4190\mathrm{J}/\mathrm{kg}.\mathrm{K}\right)\left({\mathrm{T}}_{\mathrm{f}}-293\mathrm{K}\right)+200×{10}^{-3}\mathrm{kg}\left(900\mathrm{J}/\mathrm{kg}.\mathrm{K}\right)\left({\mathrm{T}}_{\mathrm{f}}-373\mathrm{K}\right)=0 \\ 209.5{\mathrm{T}}_{\mathrm{f}}-61383.5+180{\mathrm{T}}_{\mathrm{f}}-67140=128523.5 \\ 389.5{\mathrm{T}}_{\mathrm{f}}=128523.5 \\ {\mathrm{T}}_{\mathrm{f}}=329.97\mathrm{K} \\ ≈330\mathrm{K} \end{gathered}$$

Hence, the value of the equilibrium temperature is 330K

Substituting the given values in equation (ii), we get the entropy change of aluminum as given:

$$\begin{gathered} ∆S=200×{10}^{-3}\mathrm{kg}\left(900\mathrm{J}/\mathrm{kg}.\mathrm{K}\right)In\left(\frac{329.97\mathrm{K}}{373\mathrm{K}}\right) \\ =-22.06\mathrm{J}/\mathrm{K} \\ =-22.1\mathrm{J}/\mathrm{K} \end{gathered}$$

Hence, the value of the entropy change is$-22.1\mathrm{J}/\mathrm{K}$

Substituting the given values in equation (ii), we get the entropy change of water as given:

$$\begin{gathered} ∆S=50×{10}^{-3}\mathrm{kg}\left(4190\mathrm{J}/\mathrm{kgK}\right)In\left(\frac{329.97\mathrm{K}}{293\mathrm{K}}\right) \\ =24.89\mathrm{J}/\mathrm{K} \\ ≈24.9\mathrm{J}/\mathrm{K} \end{gathered}$$

Hence, the value of the entropy change is 24.9 J/K

Change in entropy of the system = Change in entropy of water + Change in entropy of aluminum.

So, the total entropy change of the aluminum-water system is given as follows:

$$\begin{gathered} ∆S=∆S_w+∆S_{al} \\ =-22.06\mathrm{J}/\mathrm{K}+24.89\mathrm{J}/\mathrm{K} \\ ≈2.8\mathrm{J}/\mathrm{K} \end{gathered}$$

Hence, the value of the entropy change of the system is 2.8J/K