Question

A body of mass m with specific heat C at temperature 500K is brought into contact with an identical body at temperature 100K. The system is isolated from the surroundings during the process. The change in entropy of the system is

• A. mC ln 5
• B. mC ln(9/5)
• C. mC ln 3
• D. mC ln(5/3)

Answer 1

The correct option is B mC ln(9/5)

Given that the two body have same mass $m$ and specific heat $C$ and initial temperatures $T_1=500K$ and $T_2=100K$

At equilibrium, let temperature be $T_0$.

Hence,

$2mC{T}_0=mC{T}_1+mC{T}_2$

$⟹T_0=\frac{T_1+T_2}{2}$

$⟹T_0=\frac{500+100}{2}K=300K$

Now, we know that $dS=\frac{dQ}{T}$

integrating both sides:-

$⟹dS=\frac{mCdT}{T}$

$\Delta S=mC\ln \frac{T_f}{T_i}$

Hence, here, entropy change of whole system:-

$\Delta S=\Delta S_1+\Delta S_2$

$\Delta S=mC\ln \frac{T_0}{T_1}+mC\ln \frac{T_0}{T_2}$

$=mC\ln \frac{300}{500}+mC\ln \frac{300}{100}$

$\Delta S=mC\ln \frac{9}{5}$

Hence, answer is option-(B).