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Sagot :
The correct question is: 10 kg of R-134a at 300 kPa fills a rigid container whose volume is 14 L. Determine the temperature and total enthalpy in the container. The container is now heated until the pressure is 600 kPa. Determine the temperature and total enthalpy when the heating is completed.
Answer: The temperature is [tex]21.6^{o}C[/tex] and total enthalpy when the heating is completed is 300 kJ.
Explanation:
Given: Mass = 10 kg
Volume = 14 L
Final pressure = 600 K
First, convert volume from L to [tex]m^{3}/kg[/tex] as follows.
[tex]v_{1} = \frac{14 \times 10^{-3}}{10}\\= 0.0014 m^{3}/kg[/tex]
According to the R-134a tables at 300 kPa and [tex]0.0014 m^{3}/kg[/tex].
[tex]h_{1}[/tex] = 54.6 kJ/kg
[tex]T_{sat}[/tex] = 0.7 C
[tex]u_{1}[/tex] = 54.1 kJ/kg
Now, at the state 2 [tex]p_{2}[/tex] = 600 kPa and [tex]v_{2} = v_{1} = 0.0014 m^{3}/kg[/tex]
This means that the final temperature at state 2 is [tex]T_{2} = T_{sat} = 21.6^{o}C[/tex]
Hence, the change in enthalpy is calculated as follows.
[tex]\Delta H = m(h_{2} - h_{1})\\= 10(84.6 - 54.6)\\= 300 kJ[/tex]
The first law is applied to transfer the heat transfer as follows.
[tex]Q = m(u_{2} - u_{1})\\= 10(83.8 - 54.1) kJ\\= 297 kJ[/tex]
Thus, we can conclude that the temperature is [tex]21.6^{o}C[/tex] and total enthalpy when the heating is completed is 300 kJ.
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