Question

calculate the change in entropy when 100 gm of water at 30℃ is heated at 50℃. Specific heat capacity of water = 4200 J/kg.K.

Answer 1

Hola mate.......here is your answer....
The entropy change is given as

∆S=2.303×C×lg( T2/T1)

molar heat capacity is given as

= heat capacity ÷ mass

= 4200 ÷ 0.1 kg

= 420 Jkg*(2)k*(-1) = C

hence , our equation becomes as =>

∆S = 2.303×420×lg(323/303)

∆S = 26.85 J{kg*(-2)}

hope u understood ^_^