Q110 / 200 · NEET 2011 (Screening) · Physics

When $1 \mathrm{~kg}$ of ice at $0^{\circ} \mathrm{C}$ melts to water at $0^{\circ} \mathrm{C}$, the resulting change in its entropy, taking latent heat of ice to be $80 \mathrm{cal} /{ }^{\circ} \mathrm{C}$, is

  • A
    $8 \times 10^4 \mathrm{cal} / \mathrm{K}$
  • B
    $80 \mathrm{cal} / \mathrm{K}$
  • Correct answer
    C
    $293 \mathrm{cal} / \mathrm{K}$
  • D
    $273 \mathrm{cal} / \mathrm{K}$

Solution

Change in entroy $$ \Delta S=\frac{m l}{273 \mathrm{~T}}=\frac{1000 \times 80}{273}=293 \mathrm{cal} \mathrm{K}^{-1} $$

Hence, the correct answer is option: 'C'.

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