Question

01) A solid that has a latent heat of fusion Ly melts at a temperature 7. a) What is the change in entropy of this substance when a mass m of the substance melts, b) Estimate the value of the change in entropy of an ice cube of mass 30g when it melts. (Latent heat of fusion of ice is 3.33 x10 J/kg.)

Answer 1

Solution -

1. The entropy change $\Delta$ S is given by

$\Delta$S =  $\Delta$Q / T

Heat Supplied to solid (  $\Delta$Q ) = mL_f and the solid will melt at constant temperature T_{m .}

Hence entropy change $\Delta$ S is given by

$\Delta$S =   mL_f /  T_m

2. Ans -

L_f = 3.33 x 10⁵ J / Kg

m = 30 g = 0.03 Kg

T_m = 273 k ( Assuming the atmospheric pressure and temperature at which ice is kept so ice will melt at 0^oC or 273 k )

$\Delta$S =   mL_f /  T_m

$\Delta$S = ( 0.03 x 3.33 x 10⁵ ) / 273

$\Delta$S = 36.593 J / K

Thank you.

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