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

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

Tardigrade Admin · Accepted Answer

Here L =80 cal / k m =1 kg =1000 g T =273 K Heat absorbed in change of phase Q=m L Δ Q =1000 × 80=8 × 104 cal S =(Δ Q/T) =(8 × 104/273) =293 cal / k

Q. When $1 k g$ of ice at $0^{\circ} C$ melts to water at $0^{\circ} C$ , the resulting change in its entropy, taking latent heat of ice to be $80 c a l /^{\circ} C$ , is

273 cal/K

$8 \times 1 0^{4} c a l / K$

80 cal/K

293 cal/K

Here
$L = 80 c a l / k$
$m = 1 k g = 1000 g$
$T = 273 K$
Heat absorbed in change of phase $Q = m L$
$Δ Q = 1000 \times 80 = 8 \times 1 0^{4} c a l$
$S = \frac{Δ Q}{T}$
$= \frac{8 \times 1 0 ^{4}}{273}$
$= 293 c a l / k$

Q. When 1kg of ice at 0∘C melts to water at 0∘C, the resulting change in its entropy, taking latent heat of ice to be 80cal/∘C, is