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When 1 kg of ice at 0^(@)C melts to wate...

When `1 kg` of ice at `0^(@)C` melts to water at `0^(@)C`, the resulting change in its entropy, taking latent heat of ice to be `80 cal//g` is

A

`8xx10^(4)cal//K`

B

`80cal//K`

C

`293 cal//K`

D

`273 cal//K`

Text Solution

Verified by Experts

The correct Answer is:
C

change in entropy
`DeltaS = (ml)/(T) = (1000xx80)/(273)=293 cal K^(-1)`.
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