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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

293 cal/K

B

273 cal/K

C

`8xx10^4` cal/K

D

80 cal/K

Text Solution

Verified by Experts

The correct Answer is:
A
Change in entropy -
`DeltaS = (mL)/(T) =(1000 xx 80)/(273) = 293" cal "k^(-1)`
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