Question

A solid body of constant heat capacity $1J/{}^{\circ }C$ is being heated by keeping it in contact with reservoirs in two ways:
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body id was brought from initial temperature of 100${}^{\circ }C$ to a final temperature of 200${}^{\circ }C$. Entropy change of the body in two cases respectively is:
A. ln2, 4ln2
B. ln2, ln2
C. ln2, 2ln2
D. 2ln2, 8ln2

Answer 1

Hint: Entropy tells you about the disorder or randomness of the particles in a body. The change in entropy is proportional to the heat given or taken by the body or to the change in temperature of the body.

Complete step by step answer:
Entropy is the randomness of the particles in a substance. More the value of the entropy, the more randomization occurs in the substance. However, we cannot say the absolute value of entropy but we can calculate the change in the entropy of a body. The change in entropy is given as $ds=\dfrac{dQ}{T}$ …..(1), where ds is a very small change in the entropy, dQ is a very small amount of heat taken or given by the body and T is the temperature of the body. Thus change in entropy is defined as the amount of heat taken or given by a body for 1${}^{\circ }C$ rise in the temperature of the body.
For a solid body, $dQ=CdT$ …….(2), where C is the heat capacity of the body.
Substitute the value of dQ from equation (1) into equation (2).
Therefore, $ds=C\dfrac{dT}{T}$.
If the body is brought to final a temperature T2 from initial temperature T1 then the change in entropy will be,
$\Delta s=\int{ds=\int\limits_{{{T}_{1}}}^{{{T}_{2}}}{C\dfrac{dT}{T}}}$
\Rightarrow \Delta s=C\int\limits_{{{T}_{1}}}^{{{T}_{2}}}{\dfrac{dT}{T}}=\left. C\ln (T) \right|_{{{T}_{1}}}^{^{{{T}_{2}}}}=C\left\\{ \ln {{T}_{2}}-\ln {{T}_{1}} \right\\}=C\ln \left( \dfrac{{{T}_{2}}}{{{T}_{1}}} \right)
$\Rightarrow \Delta s=C\ln \left( \dfrac{{{T}_{2}}}{{{T}_{1}}} \right)$ ……..(3)
As you can see in the equation (3), the change in entropy only depends on the initial temperature and the final temperature of the body. It does not depend on the path taken to reach the final temperature from the initial temperature. It does not matter how the body is heated. That is why change in entropy is called path independent property. Since, in the both the cases the initial and the final temperatures of the body is same, the change in entropy will be same i.e. $\Delta s=C\ln \left( \dfrac{{{T}_{2}}}{{{T}_{1}}} \right)=1.\ln \left( \dfrac{200}{100} \right)=\ln 2$
Hence, the correct option is (B).

Note: In all the formulae in thermodynamics that involve temperature, the value of the temperature must be substituted in Kelvin and not in degree Celsius. Using the Celsius scale, sometimes give undefined values. For example, if the temperature of a body changes from
0${}^{\circ }C$ to 100${}^{\circ }C$ and we substitute the values of the temperatures in degree Celsius then $\Delta s=C\ln \left( \dfrac{{{T}_{2}}}{{{T}_{1}}} \right)=1.\ln \left( \dfrac{100}{0} \right)$ , which is an undefined value.