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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

273 cal/K

B

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

C

80 cal/K

D

293 cal/K

Text Solution

Verified by Experts

The correct Answer is:
D
Heat required to melt 1 kg ice at `0^(@)C` to water at `0^(@)C`
`Q=M_()L_()=(1"kg")(80"cal/g")`
`=8xx10^(4)` cal
`DeltaS=Q/(T)=(8xx10^(4)"cal")/(273K)=293" cal/K"`
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