Question

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/°C, is(a) 273 cal/K(b) 8 × 10⁴ cal/K(c) 80 cal/K (d) 293 cal/K

Answer 1

1kg=1000g

Water at 0C to ice at 0C

Q=mL

=1000×80

=8×10^4cal/K

The ans is (b)

Answer 2

right option -d

293 cal/k

change in entropy

∆s= ml/ T=1000*80/273=293cal/k^-1

I hope this answer is fully help u

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