Answer:
Q = -68.859 kJ
Explanation:
given details
mass ![co_2 = 1 kg](https://tex.z-dn.net/?f=co_2%20%3D%201%20kg)
initial pressure P_1 = 104 kPa
Temperature T_1 = 25 Degree C = 25+ 273 K = 298 K
final pressure P_2 = 1068 kPa
Temperature T_2 = 311 Degree C = 311+ 273 K = 584 K
we know that
molecular mass of ![co_2 = 44](https://tex.z-dn.net/?f=co_2%20%3D%2044)
R = 8.314/44 = 0.189 kJ/kg K
c_v = 0.657 kJ/kgK
from ideal gas equation
PV =mRT
![V_1 = \frac{m RT_1}{P_1}](https://tex.z-dn.net/?f=V_1%20%3D%20%5Cfrac%7Bm%20RT_1%7D%7BP_1%7D)
![=\frac{1*0.189*298}{104}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%2A0.189%2A298%7D%7B104%7D)
![V_1 = 0.5415 m3](https://tex.z-dn.net/?f=V_1%20%3D%200.5415%20m3)
![V_2 = \frac{m RT_2}{P_2}](https://tex.z-dn.net/?f=V_2%20%3D%20%5Cfrac%7Bm%20RT_2%7D%7BP_2%7D)
![=\frac{1*0.189*584}{1068}](https://tex.z-dn.net/?f=%20%3D%5Cfrac%7B1%2A0.189%2A584%7D%7B1068%7D)
![V_1 = 0.1033 m3](https://tex.z-dn.net/?f=V_1%20%3D%200.1033%20m3)
WORK DONE
![W =P_{avg}*{V_2-V_1}](https://tex.z-dn.net/?f=W%20%3DP_%7Bavg%7D%2A%7BV_2-V_1%7D)
w = 586*(0.1033 -0.514)
W =256.76 kJ
INTERNAL ENERGY IS
![\Delta U = m *c_v*{V_2-V_1}](https://tex.z-dn.net/?f=%5CDelta%20U%20%20%3D%20m%20%2Ac_v%2A%7BV_2-V_1%7D)
![\Delta U = 1*0.657*(584-298)](https://tex.z-dn.net/?f=%5CDelta%20U%20%20%3D%201%2A0.657%2A%28584-298%29)
![\Delta U =187.902 kJ](https://tex.z-dn.net/?f=%5CDelta%20U%20%20%3D187.902%20kJ)
HEAT TRANSFER
![Q = \Delta U +W](https://tex.z-dn.net/?f=Q%20%3D%20%5CDelta%20U%20%20%2BW)
= 187.902 +(-256.46)
Q = -68.859 kJ
$2.
Both tickets cost $1.50
$1.50 x 2 = $3
$5 - $3 = $2
Answer:
fluid nozzle that is too large