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 120 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 $\frac{KJ}{kg K}$

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 $\frac{KJ}{kg}$

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

⇒ ΔS = $\frac{212.32}{250.5}$

⇒ ΔS = 0.847 $\frac{KJ}{kg K}$

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