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One mole of ice is melted at 0° C and then is heated to 100° C, and then vaporized at 100° C. What is the difference in entropies of the steam and ice? The heats of vaporization and fusion are $540 cal g^{- 1}$ and $80 cal g^{- 1}$ respectively. Use the average heat capacity of liquid water as $1 cal g^{- 1} {degree}^{- 1} :$

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$18 (\frac{80}{373} + \frac{540}{273} + \ln \frac{373}{273})$

$18 (\frac{80}{273} + \frac{540}{373} + \ln \frac{373}{273})$

$(\frac{80}{273} + \frac{540}{373} + \ln \frac{273}{373})$

$(\frac{80}{273} + \frac{540}{373} + 100)$

answer is B.

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Detailed Solution

$Ice ⟶ water ⟶ water ⟶ steam$

$(0^{\circ} C) (0^{\circ} C) (100^{\circ} C) (100^{\circ} C)$

$Δ S_{Total} = 18 \times \frac{80}{273} + 18 \times 1 \times \ln (\frac{373}{273}) + 18 \times \frac{540}{373}$

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