Answer:
The specific heat capacity is q_{L}=126.12kJ/kg
The efficiency of the temperature is n_{TH}=0.67
Explanation:
The p-v diagram illustration is in the attachment
T_{H} means high temperature
T_{L} means low temperature
The energy equation :
[tex]q_{h}[/tex] = R* [tex]T_{h}[/tex] in([tex]V_{2}[/tex]/[tex]V_{1}[/tex])
[tex]=0.287 * 1200 ln(3)[/tex]
[tex] =0.287*1318.33[/tex]
[tex] =378.36kJ/kg[/tex]
The specific heat capacity:
[tex]q_{L}[/tex]=q_{h}*(T_{L}/T_{H})
q_{L}=378.36 * (400/1200)
q_{L}=378.36 * 0.333
q_{L}=126.12kJ/kg
The efficiency of the temperature will be:
[tex]n_{TH}[/tex]=1 - ([tex]T_{L}[/tex]/[tex]T_{H}[/tex])
n_{TH}=1-(400/1200)
n_{TH}=1-0.333
n_{TH}=0.67