Question

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

Answer 1

Hint: Entropy change of body with respect to temperature can be given by the formula, $\Delta S{}^\circ =\int\limits_{A}^{B}{\dfrac{dQ}{T}}$. Now, the change in heat energy can be given by the formula, $Q=mc\Delta T$. We will consider this change in heat energy for small changes in temperature and based on that we will find the answer for both the cases.

Formula used: $\Delta S{}^\circ =\int\limits_{A}^{B}{\dfrac{dQ}{T}}$, $Q=mc\Delta T$

Complete step by step answer:
In question we are given that a solid body of constant heat capacity $1\ J{}^\circ C$, is being heated by the reservoirs in two ways. And it is given that in both cases the same amount of heat is supplied by the reservoir and the change in temperature is from $100\ J{}^\circ C$ to $200\ J{}^\circ C$. And we are asked to find the change in entropy in both cases. So first of all formula of change in entropy can be given as,
$\Delta S{}^\circ =\int\limits_{A}^{B}{\dfrac{dQ}{T}}$ …………………….(i)
Where, Q is change in heat energy and T is temperature in kelvin.
Now, ${{T}_{A}}=100{}^\circ C$, and ${{T}_{B}}=200{}^\circ C$
Now, we know that conversion of temperature from degree Celsius to can Kelvin can be done by adding 273 to each temperature, which can be seen mathematically as,
${{T}_{A}}=100+273=373k$
${{T}_{B}}=200+273=473k$
Now, substituting these values in (i) we will get,
$\Delta S{}^\circ =\int\limits_{{{T}_{B}}}^{{{T}_{A}}}{\dfrac{dQ}{T}}$
Now, change in heat energy can be given by the formula,
$Q=mc\Delta T$
Where, Q is change in heat energy, m is mass, c is heat capacity constant and T is temperature in ${}^\circ C$.
On considering the changes as small, the above expression can be given as,
$dQ=mcdT$
Now, as the change in temperature is in degree Celsius, it can be shown as,
$T'=T-273$ ………(ii)
And $dT'=dT$
On substituting these values in expression (i) we will get,
$\Delta S{}^\circ =\int\limits_{{{T}_{B}}}^{{{T}_{A}}}{\dfrac{cdT'}{T-273}}$
Now, here ${{T}_{A}}=373-273=100{}^\circ C$ and ${{T}_{A}}=473-273=200{}^\circ C$
On substituting these values in expression, we will get,
$\Delta S{}^\circ =\int\limits_{100}^{200}{\dfrac{cdT'}{T'+273}}$
Now, integration of $\int{\dfrac{dx}{x}=\ln x}$, so applying this rule we will get,
$\Delta S{}^\circ =\int\limits_{100}^{200}{\dfrac{cdT'}{T'+273}}\Rightarrow \Delta S{}^\circ =\ln \left[ T+273 \right]_{100}^{200}$
$\Rightarrow \Delta S{}^\circ =\ln \left[ \left( 200+273 \right)-\left( 100+273 \right) \right]$
Now, we know that $\ln \left( b-a \right)=\ln \dfrac{b}{a}$, so applying these values we will get,
$\Rightarrow \Delta S{}^\circ =\ln \left( \dfrac{473}{373} \right)$
$\Rightarrow \Delta S{}^\circ =\ln 2$
Now, as the heat transfer is the same entropy change will also be the same in both the cases. So, the answer is $\ln 2,\ \ln 2$.
Hence, option (b) is the correct answer.

Note: As, we know the fact that if the change in heat energy is same in both cases then entropy change will also be same as change in entropy is directly proportional to change in entropy, which can be shown mathematically as, $\Delta S{}^\circ \propto \Delta Q$. So, this can be solved by trial and error method by looking at the values in each option and then selecting the option which has similar values we will get an answer. So, this can be considered as an alternative method to the above solution.