# The latent heat of ice is 80 Cal/gm. The change in entropy when 10 gram of ice at 0 Â°C is converted into water of same temperature i

given, latent heat of ice=80cal/gm

i.e., for 1 gm of ice, hea needed is 80 cal/gm, so heat needed for 10 gm of ice=80 x 10=800 cal

Temperature=0oC=273 K

we know that change in entropy is given by:-

$\Delta S=\frac{\Delta Q}{\Delta T}=\frac{800}{273}=2.93cal/K$

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