Calculate the entropy change when 1 mol of ice 0- C is converted into water at 0- C- The change in heat is 1436 cal per mol-

Do you remember, what is mathematical definition of entropy? Yes, it was dS = dqT (reversible process) hence, Δ S =q_revT = 1436273 = 5.26 caldeg^ 1 mol^ ...

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Standard XII

Physics

2nd Law of Thermodynamics

Calculate the...

Question

Calculate the entropy change when 1 mol of ice $0^{\circ} C$ is converted into water at $0^{\circ} C$ . The change in heat is 1436 cal per mol.

8.56 c a l d e g^{- 1} m o l^{- 1}

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9.96 c a l d e g^{- 1} m o l^{- 1}

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$5.26 c a l d e g^{- 1} m o l^{- 1}$

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Solution

The correct option is C
$5.26 c a l d e g^{- 1} m o l^{- 1}$

Do you remember, what is mathematical definition of entropy?
Yes, it was
$d S = \frac{d q}{T}$ (reversible process)
hence, $Δ S = \frac{q_{r e v}}{T}$
= $\frac{1436}{273}$ $= 5.26 c a l d e g^{- 1} m o l^{- 1}$

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Similar questions

Q. What is the entropy change (in

{J K}^{- 1} {m o l}^{- 1}

) when one mole of ice is converted into water at

0^{o} C

Calculate the entropy change accompanying the following change of state.

$H_{2} O (s, - 10^{\circ} C, 1 a t m) \to H_{2} O (l, 10^{\circ} C, 1 a t m)$
$C_{P}$ for ice = $9 c a l d e g^{- 1} m o l^{- 1}$

$C_{P}$ for $H_{2} O$ = $18 c a l d e g^{- 1} m o l^{- 1}$

Latent heat of fusion of ice = $1440 c a l m o l^{- 1}$ at $0^{\circ} C$

Q. Calculate the change in entropy when

1 g

of ice at

0^{\circ}

C is heated to form water at

40^{\circ} C

Q. When 1 kg of ice

0^{\circ} C

melts to water

0^{\circ} C

, the resulting change in its entropy , taking latent heat of ice to be 80 Cal/gm is-

Q. At

0^{o} C

, ice and water are in equilibrium and enthalpy change for the process

H_{2} O (s) ⟶ H_{2} O (l)

6 k J {m o l}^{- 1}

. Calculate the entropy change for the conversion of ice into liquid water.

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Calculate the entropy change when 1 mol of ice 0∘C is converted into water at 0∘C. The change in heat is 1436 cal per mol.

The correct option is C 5.26 caldeg−1mol−1 Do you remember, what is mathematical definition of entropy? Yes, it was dS=dqT (reversible process) hence, ΔS=qrevT = 1436273 =5.26 caldeg−1mol−1

Calculate the entropy change when 1 mol of ice $0^{\circ} C$ is converted into water at $0^{\circ} C$ . The change in heat is 1436 cal per mol.

The correct option is C
$5.26 c a l d e g^{- 1} m o l^{- 1}$

Do you remember, what is mathematical definition of entropy?
Yes, it was
$d S = \frac{d q}{T}$ (reversible process)
hence, $Δ S = \frac{q_{r e v}}{T}$
= $\frac{1436}{273}$ $= 5.26 c a l d e g^{- 1} m o l^{- 1}$