Calculate the change in entropy when a 279 g ice cube at 12

Calculate the change in entropy when a 27.9 g ice cube at -12 degrees celsius is heated to yield steam at 115 degrees celsius.

Mass m = 27.9 g

Initial temprature T = -12 ^o C = -12 + 273 = 261 K

Temprature of fusion of ice T \' = 0 ^o C = 273 K

Temprature of stream T \" = 100 ^oC = 100 + 273 = 373 K

Final temprature T \'\" = 115 ^o C = 115 + 273 = 388 K

Change in entropy dS = mC ln(T \'/ T) +(mL/T \')+mC \' ln(T \" / T \') + (mL \' /T \") + mC \" ln ( T\"\'/T\")

= m{ C ln(T \'/ T) +(L/T \')+C \' ln(T \" / T \') + (L \' /T \") + C \" ln ( T\"\'/T\") }

Where C = Specific heat of ice = 2.108 J / g K

C \' = Specific heat of water = 4.19 J / g K

C \" = Specific heat of steam = 1.996 J / g K

L = Latent heat of fusion = 334 J / g

L \' = latent heat of vaporisation of water = 2256 J / g

Substitute values you get ,

dS = 27.9 {[2.108 ln(273/261)]+(334/273)+[4.19 ln(373/273)]+(2256/373)+[1.996 ln(388/373)]}

= 27.9 {0.0947575 + 1.22344 + 1.30772 + 6.0482 + 0.07869}

= 27.9 x 8.7632

= 244.5 J / K