Question

what is the change in entropy when 50g of hot water at 80oC is poured into 100g of cold water at 10oC in an insulated container. Assume constant pressure, use Cp,m=75.3j/(molk)

Answer 1

Step 1

Find out the final temperature (equilibrium Temperature) using following equation

- ( mol of hot water x Cpm x (Tfinal – Tinitial) = mol of cold water x Cpm x (Tfinal -T initial)

Lets plug given values to get the final T.

- ( 50/18.0148 x 75.3 J/ K mol x ( Tf-80 ) = (100/18.0148) x 75.3 J / K mol x ( Tf – 10 deg C)

Lets find Tf

Tf = 33.33 deg C

Convert Tf into Kelvin

= 33.33 + 273.15 = 306.15 K

Also convert initial T of both into K

Initial T of hot water = 80+273.15 = 353.15 K

Initial T of cold water = 10 +273.15 = 283.15 K

Delta S = n Cp,m x ln ( Tf/Ti)

Delta S = Delta S of hot water + delta S of cold water

= (50/18.0148) x75.3 x ln ( 306.15/353.15) + ( 100/18.0148) x 75.3 xln ( 306.15/283.15)

= 2.79 J / K

Or 2.80 J /K