Question

In an experiment, 200 gof aluminum (with a specific heat of 900J/kg.K) at 100$°$Cis mixed with 50.0gof water at 20.0$°$C, with the mixture thermally isolated.(a)What is the equilibrium temperature?(b)What is the entropy changes of the aluminum, (c)What is the entropy changes of the water, and (d)What is the entropy changes of the aluminum – water system?

Short answer

a) The equilibrium temperature of aluminum-water system is 330K.

b) Change in entropy of the aluminum is $-22.1\mathrm{J}/\mathrm{K}$.

c) Change in entropy of the water is 24.9 J/K

d) Change in entropy of the aluminum-water system is 2.8 J/K

Step-by-step solution

The given data

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

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

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,$c_{al}=4190\mathrm{J}/\mathrm{kg}.\mathrm{K}$

Understanding the concept of equilibrium and entropy change

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,$Q=mc∆T$ …(i)

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

(a) Calculation of the equilibrium temperature of aluminum-water system

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

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

Hence, the value of the equilibrium temperature is 330K

(b) Calculation of the entropy change of aluminum

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

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

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

(c) Calculation of the entropy change of water

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

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

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

(d) Calculation of the entropy change of aluminum-water system

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:

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

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