JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Chemical Thermodynamics

Two blocks of the same metal having 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,

Δ S

, for this process is:

Answer: (A)

Solution

Equal masses of the same metal settle at the common final temperature

T_{f} = \frac{T _{1} + T _{2}}{2}

. Each block's entropy change at constant pressure is

C_{p} ln \frac{T _{f}}{T _{i}}

, the hotter losing and the cooler gaining, so the total is

Δ S = C_{p} ln \frac{T _{f}}{T _{1}} + C_{p} ln \frac{T _{f}}{T _{2}} = C_{p} ln \frac{T _{f}^{2}}{T _{1} T _{2}}

. Putting in

T_{f} = \frac{T _{1} + T _{2}}{2}

gives

Δ S = C_{p} ln \frac{( T _{1} + T _{2} ) ^{2}}{4 T _{1} T _{2}}

JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Single correct option