Solution Found!
Using Entropy Data and ConceptsPropane undergoes a process
Chapter 6, Problem 10P(choose chapter or problem)
Propane undergoes a process from state 1, where \(p_{1}=1.4\) MPa, \(T_{1}=60^{\circ} \mathrm{C}\), to state 2, where \(p_{2}=1.0\) MPa, during which the change in specific entropy is \(s_{2}-s_{1}=-0.035 \mathrm{\ kJ} / \mathrm{kg} \cdot \mathrm{K}\). At state 2, determine the temperature, in \({ }^{\circ} \mathrm{C}\), and the specific enthalpy, in kJ/kg.
Questions & Answers
QUESTION:
Propane undergoes a process from state 1, where \(p_{1}=1.4\) MPa, \(T_{1}=60^{\circ} \mathrm{C}\), to state 2, where \(p_{2}=1.0\) MPa, during which the change in specific entropy is \(s_{2}-s_{1}=-0.035 \mathrm{\ kJ} / \mathrm{kg} \cdot \mathrm{K}\). At state 2, determine the temperature, in \({ }^{\circ} \mathrm{C}\), and the specific enthalpy, in kJ/kg.
ANSWER:Step 1 of 2
We are required to calculate the temperature and specific enthalpy.