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Two blocks of the same metal having the same mass and at temperature ${{T}_{1}}$ and ${{T}_{2}}$ respectively are brought in contact with each other and allowed to attain thermal equilibrium at constant pressure. The change in entropy, $\Delta S$, for this process is:
(a) $2{{C}_{P}}\ln \left( \dfrac{{{T}_{1}}+{{T}_{2}}}{4{{T}_{1}}{{T}_{2}}} \right)$
(b) $2{{C}_{P}}\ln \left( \dfrac{{{({{T}_{1}}+{{T}_{2}})}^{\dfrac{1}{2}}}}{{{T}_{1}}{{T}_{2}}} \right)$
(c) ${{C}_{P}}\ln \left( \dfrac{{{({{T}_{1}}+{{T}_{2}})}^{2}}}{4{{T}_{1}}{{T}_{2}}} \right)$
(d) $2{{C}_{P}}\ln \left( \dfrac{{{T}_{1}}+{{T}_{2}}}{2{{T}_{1}}{{T}_{2}}} \right)$


Answer
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Hint: Entropy tells us about the disorder or uncertainty in the system and we know that the two metal blocks are in contact with each other and have the temperatures as ${{T}_{1}}$ and ${{T}_{2}}$ respectively and we can calculate the total entropy of the system by using the formula as $\Delta S$ = $n{{C}_{p}}\int{\dfrac{dT}{T}}$ here, ${{C}_{P}}$ is the specific heat at constant pressure and dT is the change in temperature.
Now, solve it.

Complete step by step answer:
First of all ,we should know what entropy is. By the term entropy we mean the degree of randomness in a system or simply we can say the uncertainty or disorder of a system.
The Entropy and the amount of heat generated are related to each other and the amount of entropy change due to the change in the amount of heat depends on the temperature. If suppose ds is the change in entropy and dq is the change in energy and T is the temperature , then;
$\Delta S$ =$\int{\dfrac{dq}{T}}$
As we know that;
dq=$n{{C}_{P}}dT$
Here, ${{C}_{P}}$ is the specific heat at constant pressure and dT is the change in temperature.
Put the value of dq in above equation, we get;
$\Delta S$ =$\int{\dfrac{n{{C}_{P}}dT}{T}}$
$\Delta S$ = $n{{C}_{p}}\int{\dfrac{dT}{T}}$--------------(1)
Now, considering the numerical as;
As we know that the two blocks are in close contact with each other and ${{T}_{1}}$is the temperature of one block and${{T}_{2}}$is the temperature of the other block (given), then the total temperature of the two blocks is;
\[{{T}_{f}}=\dfrac{{{T}_{1}}+{{T}_{2}}}{2}\] --------(2)
 Then,
The entropy of first block of metal can be found by using the equation (1) as;
$\Delta {{S}_{1}}=n{{C}_{P}}\int\limits_{{{T}_{1}}}^{{{T}_{f}}}{\dfrac{dT}{T}}$
=$n{{C}_{P}}\ln \dfrac{{{T}_{f}}}{{{T}_{1}}}$ ---------------(3)
Similarly, the entropy of second metal block is;
$\Delta {{S}_{2}}=n{{C}_{P}}\int\limits_{{{T}_{2}}}^{{{T}_{f}}}{\dfrac{dT}{T}}$
=$n{{C}_{P}}\ln \dfrac{{{T}_{f}}}{{{T}_{2}}}$ ----------------(4)
The total change in entropy can be calculated as;
$\Delta {{S}_{total}}=\Delta {{S}_{1}}+\Delta {{S}_{2}}$
Put the values of equation (3) and (4)in it, we get;
$\Delta {{S}_{total}}=n{{C}_{P}}\ln \dfrac{{{T}_{f}}}{{{T}_{1}}}+n{{C}_{P}}\ln \dfrac{{{T}_{f}}}{{{T}_{2}}}$
=$n{{C}_{P}}\ln \left( \dfrac{{{T}_{f}}}{{{T}_{1}}}+\dfrac{{{T}_{f}}}{{{T}_{2}}} \right)$
$\Delta {{S}_{total}}$ =$n{{C}_{P}}\ln \left( \dfrac{{{({{T}_{f}})}^{2}}}{{{T}_{1}}{{T}_{2}}} \right)$
Put the value of equation(2) in it; we get
$\Delta {{S}_{total}}$ =$n{{C}_{P}}\ln \left( \dfrac{{{({{T}_{1}}+{{T}_{2}})}^{2}}}{4{{T}_{1}}{{T}_{2}}} \right)$
Thus, for the two blocks of the same metal having the same mass and temperature as ${{T}_{1}}$ and ${{T}_{2}}$ . the total change in entropy,$\Delta S$ is: $n{{C}_{P}}\ln \left( \dfrac{{{({{T}_{1}}+{{T}_{2}})}^{2}}}{4{{T}_{1}}{{T}_{2}}} \right)$.

Hence, option(c) is correct.

Note: Entropy is used to describe the behaviour of a system in terms of thermodynamic properties such as temperature, pressure, entropy, and heat capacity etc. the entropy of solids is always more than the entropy of gases because in solids the molecules are closely packed as compared to the gases. Entropy is related to the second law of thermodynamics according to which the entropy of the universe always increases in a spontaneous process.