91Ó°ÊÓ

a) Mass of copper,$m_c=50.0g$

b) Temperature of copper at an insulating end,$T_c=400k$

c) Mass of lead,$m_l=100g$

d) Temperature of lead,$T_l=200k$

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

Formulae:

The amount of heat absorbed by the body,$Q=mc∆T$ . …(i)

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

At equilibrium, the total heat transferred by the system is zero. So,$Q_c+Q_l=0$

Now, using equation (i), we get that

$m_c{c}_c∆T_c+m_l{c}_l∆T_l=0$

$c_c$is specific heat of copper = $386J/kgK$

$c_l$is specific heat of lead = $128J/kgk$

Let equilibrium temperature be $T_f.$

Then, the above equation becomes

$$\begin{gathered} m_c{c}_c\left(T_f-T_c\right)+m_l{c}_l\left(T_f-T_l\right)=0 \\ 50×{10}^{-3}kg\left(386J/kg\right)\left(T_f-400k\right)+100×{10}^{-3}kg\left(128J/kg\right)\left(T_f-400k\right)=0 \\ 19300×{10}^{-3}{T}_f-7720000×{10}^{-3}+12800×{10}^{-3}{T}_f-2560000×{10}^{-3}=0 \\ 32100T_f=1028000 \\ {T}_f=320.249k \\ ~320.25k \end{gathered}$$

Hence, the value of equilibrium temperature is 320.25K

The two-block system is thermally insulated. So, the total internal energy of the system remains constant. Therefore, the change in the internal energy of the system between initial state and the equilibrium state is zero.

Change in entropy of the system = Change in entropy of the copper block + Change in entropy of the lead block

Change in the entropy using equation (ii) and the given data is given as follows:

$$\begin{gathered} ∆S=∆S_c+∆S_l \\ =m_c{c}_c\ln \left(\frac{T_f}{T_c}\right)+m_l{c}_l\ln \left(\frac{T_f}{T_l}\right) \\ =50×{10}^{-3}kg\left(386J/kgk\right)\ln \left(\frac{320.249k}{400k}\right)+100×{10}^{-3}kg\left(128J/kgk\right)\ln \left(\frac{320.249k}{200k}\right) \\ =1734.34×{10}^{-3}J/k \\ ≈1.73J/K \end{gathered}$$

Hence, the value of the entropy change of the system is 1.73 J/K