Question

Using the relation dS =

(δQ/T)

int rev for the

definition of entropy,

calculate the change in

the specific entropy of

R-134a as it is heated at

a constant pressure of

240 kPa from a

saturated liquid to a

saturated vapor. Use

the tables for R-134a.

​

Answer 1

Answer:

The value of specific entropy = 0.847 \frac{KJ}{kg K} kgK KJ

Explanation:

Constant pressure = 120 K pa

From the property tables of R - 134 a at P = 120 K pa

Temperature T = - 22.5 ° c

Value of heat transfer = 212.32 \frac{KJ}{kg}kgKJ

Thus entropy change ΔS = \frac{dq}{T}Tdq

⇒ ΔS = \frac{212.32}{250.5}250.5212.32

⇒ ΔS = 0.847 \frac{KJ}{kg K}kgKKJ

Therefore the value of specific entropy = 0.847 \frac{KJ}{kg K}kgKKJ