Answer:
heat transfer is 4.6067 kJ
Explanation:
given data
P1= 2 bar = 2 × [tex]10^{5}[/tex] N/m²
P2= 8 bar = 8 × [tex]10^{5}[/tex] N/m²
V2= 0.02 m³
m = 0.2 kg
[tex]PV^{2=1.3}[/tex] = constant
V(1 to 2) = 50 kJ/kg
U(1 to 2) = m V(1 to 2) = 0.2 × 50 = 10 kJ
solution
[tex]P_1 V_1^{1.3} = P_2 V_2^{1.3}[/tex]
[tex]2*10^5 V_1^{1.3} = 8*10^5 * 0.02^{1.3}[/tex]
v1 = 0.05809 m³
and
by 1st law of thermodynamics for closed system is
Q - W = ΔU
so calculate the work expansion that is
W = [tex]\frac{P_1V_1 - P_2 V_2}{n-1}[/tex]
W = [tex]\frac{2*10^5 * 0.05809 - 8*10^5 * 0.02}{1.3-1}[/tex]
W = -14.6067 kJ
so heat transfer from 1st law of thermodynamic is
Q = U + W
Q = 10 + ( - 14.06067 )
Q = -4.6067kJ
so
heat transfer is 4.6067 kJ
