Question

4.5kg of water at 25?C is mixed with 4.5kg of water at 58?C in a well-insulated container.Estimate the net change in entropy of the system.

Answer 1

Mass of water m = 4.5 Kg

Specific heat of water S = 4186 J/Kg K

                                     = 1 K cal/Kg K

                                     = 1000 cal/Kg K

Average temperature t = (25 + 58)/2 = 41.5⁰C

Heat Q = mS ?t

            = (4.5 Kg)(1000 cal/Kg K)(58 - 25)

            = 148500 cal

we need to solve for two T's the hot water T and the cold water T

       T for hot water Th = (58 + 41.5)/2 = 49.75 + 273 = 322.75 K

       T for cold water Tc = (41.5 + 25)/2 = 33.25 + 273 = 306.25 K

  

Now change in entropy ?S = - (148500 cal/322.75 K) + (148500 cal/306.25 K)

                                         = - -460.108 +484.89

                                         = + 24.79 cal/K