Answer:
-50.005 KJ
Explanation:
Mass flow rate = 0.147 KJ per kg
mass= 10 kg
Δh= 50 m
Δv= 15 m/s
W= 10×0.147= 1.47 KJ
Δu= -5 kJ/kg
ΔKE + ΔPE+ ΔU= Q-W
0.5×m×(30^2- 15^2)+ mgΔh+mΔu= Q-W
Q= W+ 0.5×m×(30^2- 15^2) +mgΔh+mΔu
= 1.47 +0.5×1/100×(30^2- 15^2)-9.7×50/1000-50
= 1.47 +3.375-4.8450-50
Q=-50.005 KJ
Answer:
50.005 kJ of heat is transferred by the system
Solution:
As per the question:
Mass of the closed system, m = 10 kg
Decrease in elevation, d = 50 m
Initial velocity, v = 15 m/s
Final velocity, v' = 30 m/s
Change in internal energy, dU = - 5 kJ/kg = - 5000(10 kg) = - 50000 J
As per the first thermodynamics' law:
dQ = dW + (dPE +dKE + dU) (1)
where
dQ = heat transfer change
dPE +dKE + dU = dE = change in the energy of the system
PE = Potential Energy
KE = Kinetic Energy
U = Internal energy
Now,
dPE = mgd = - [tex]10\times 9.7\times 50 = 4850 J = - 4.850 kJ[/tex]
(Since, the elevation decreases and hence PE decreases)
dKE = [tex]\frac{1}{2}m(v'^{2} - v^{2}) = \frac{1}{2}\times 10(30^{2} - 15^{2}) = 3375 J = 3.375 kJ[/tex]
Work done, dW = [tex]0.147\times 10^{3}\times 10 = 1470 J[/tex]
Therefore, using the respective energy values in eqn (1):
[tex]dQ = 1470 + (- 4850 + 3375 - 50000) = - 50005 J = - 50.005 kJ[/tex]
