Question

A frictionless piston-cylinder device contains a saturated liquid and water vapor mixture at 110 ℃. During a constant pressure process, 546 kJ of heat is transferred to the surrounding air at 15 ℃. As a result, the vapor part inside the cylinder condenses. Determine a) the entropy change of the water and b) the generation of total entropy during this heat transfer process.

Answer 1

Explanation:

(a)   It is given that heat transferred is -546 kJ (as heat is releasing) and temperature is 110 degree celsius.

Hence, the temperature in kelvin will be as follows.

                     (110 + 273) K

                   = 383 K

As it is known that relation between entropy change and heat is as follows.

                     S = $\frac{Q}{T}$

Putting the given values in the above equation as follows.

                   S = $\frac{Q}{T}$

                      = $\frac{546 kJ}{383 K}$  

                      = -1.4255 kJ/K

As temperature changes from $110 ^{o}C$ to $15 ^{o}C$. So, (15 + 273) K = 288 K. Hence, change in entropy will be calculated as follows.

                $\Delta S = \frac{Q}{T}$

                             = $\frac{546 kJ}{288 K}$

                             = 1.895 kJ/K

Therefore, entropy change of water will be 1.895 kJ/K.

(b)    As total entropy generated will be the sum of both the entropies as follows.

              Total entropy = (-1.4255 kJ/K + 1.895 kJ/K)

                                     = 0.4695 kJ/K

Thus, total entropy during this heat transfer process is 0.4695 kJ/K.