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JEE Main & Advanced JEE Main Paper (Held On 11-Jan-2019 Morning)

  • question_answer
    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, \[\Delta S\], for this process is [JEE Main Online Paper (Held On 11-Jan-2019 Morning]

    A) \[2{{C}_{p}}\ln \left[ \frac{{{T}_{1}}+{{T}_{2}}}{2{{T}_{1}}{{T}_{2}}} \right]\]                   

    B) \[{{C}_{p}}\ln \left[ \frac{{{({{T}_{1}}+{{T}_{2}})}^{2}}}{4{{T}_{1}}{{T}_{2}}} \right]\]

    C) \[2{{C}_{p}}\ln \left[ \frac{({{T}_{1}}+{{T}_{2}})}{4{{T}_{1}}{{T}_{2}}} \right]\]     

    D)   \[2{{C}_{p}}\ln \left[ \frac{{{({{T}_{1}}+{{T}_{2}})}^{\frac{1}{2}}}}{{{T}_{1}}{{T}_{2}}} \right]\]

    Correct Answer: B

    Solution :

    \[{{T}_{final}}=\frac{{{T}_{1}}+{{T}_{2}}}{2}\] \[\Delta {{S}_{1}}={{C}_{p}}\ln \frac{{{T}_{f}}}{{{T}_{1}}},\Delta {{S}_{2}}={{C}_{p}}\ln \frac{{{T}_{f}}}{{{T}_{2}}}\] \[\Delta S=\Delta {{S}_{1}}+\Delta {{S}_{2}}={{C}_{p}}\ln \frac{T_{f}^{2}}{{{T}_{1}}{{T}_{2}}}={{C}_{p}}\ln \left[ \frac{{{({{T}_{1}}+{{T}_{2}})}^{2}}}{4{{T}_{1}}{{T}_{2}}} \right]\]


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