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  • Find the change in the entropy in the following process of ice at <img alt="0^{\circ} \ma

Find the change in the entropy in the following process 80 \mathrm{~g} of ice at 0^{\circ} \mathrm{C} melts when dropped in a bucket of water at 50^{\circ} \mathrm{C}. (Assume temperature of water does not change)

Option: 1

\mathrm{-5.4 \, calK ^{-1}}


Option: 2

\mathrm{+4.5 \, calK1 ^{-1}}


Option: 3

\mathrm{+5.4 \, calK ^{-1}}


Option: 4

\mathrm{-5.4 \, calK ^{-1}}


Answers (1)

best_answer

Gain of entropy of ice

\mathrm{S_{1} =\frac{\Delta Q}{T}=\frac{m L}{T}}
\mathrm{=\frac{80 \times 100}{(0+273)}=\frac{8 \times 10^{3}}{273} \mathrm{cal} \mathrm{K}^{-1}}

Loss of entropy of water
\mathrm{S_{2} =-\frac{\Delta Q}{T}=-\frac{m L}{T} }
\mathrm{ =\frac{80 \times 100}{(273+50)}=\frac{8 \times 10^{3}}{323} \mathrm{cal} \mathrm{K}^{-1} }

Total change of entropy
\mathrm{S_{1}+S_{2} =\frac{8 \times 10^{3}}{273}-\frac{8 \times 10^{3}}{323} }
                  \mathrm{=+4.5 \mathrm{calK}^{-1} }

Posted by

Irshad Anwar

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