Answer:
[tex]final-temperature = T_{f} = 252.51K[/tex]
Explanation:
we can solve this problem by using the first law of thermodynamics.
[tex]\Delta U= Q-W[/tex]
Q= heat added
U= internal energy
W= work done by system
[tex]E_{final}= E_{initial}[/tex]
[tex]C_{v} (N_{2}) C_{v}(He) T_{f} =C_{v}( N_{2}) T_{N_{2} } C_{v} (He) T_{He}[/tex] (1)
[tex]C_{v}(N_{2})=1.04\frac{KJ}{Kg K}[/tex]
[tex]C_{v}(He)=5.193\frac{KJ}{Kg K}[/tex]
now
From equation 1
[tex]T_{f}=\frac{C_{v}( N_{2}) T_{N_{2} } C_{v} (He) T_{He}}{C_{v} (N_{2}) C_{v}(He)}[/tex]
[tex]T_{f} = \frac{315\times1.04+5.193\times240}{1.04+5.193}[/tex]
[tex]T_{f} = 252.51K[/tex]