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An unknown gas (k=1.4, c v= 0.716 kJ/kg.K, c_p=1 kJ/kg.K, R = 0287 kJ/kg K) is trapped in a 1 m^3 piston-cylinder device at 1000 KPa and 1000 K. It then undergoes an isothermal (constant-temp) process in which 696 kJ of boundary work is delivered (positive work transfer). Determine the heat transfer. (Use the PG model).

Respuesta :

Answer:

So heat transfer is 696 kJ

Explanation:

Given:

K = 1.4

[tex]C_{v}[/tex] = 0.716 kJ/kg

[tex]C_{p}[/tex] = 1 kJ/kg

R = 0.287 kJ/kg

V = 1 [tex]m^{3}[/tex]

P = 1000 kPa

T = 1000 K

Work delivered, δW = 696 kJ

It is isothermal process, so the initial and final temperature are same, that is T₁ = T₂ and the internal energy is zero (dU =0)

Therefore from 1st law of thermodynamcis,

δQ = dU + δW

     = 0 + 696

     = 696 kJ

So heat transfer is 696 kJ

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