Respuesta :
There will be no heat transmission, and the system's work is 25.27 Kj.
The calculation for total heat transfer and work production:
Provided,
The mass of r-134a = 0.7 kg
Initial pressure of the system, P₁ = 800 kpa
Initial temperature of r-134a, T₁ = 50°C = 273+ 50 = 323 K
Final pressure of the system, P₂ = 140 kpa
From the superheated refrigerant r-134a
P₁ = 800 kpa & T₁ = 50°C
the internal energy, u₁ = 263.87 Kj/kg &
the entropy, s₁ = 0.9803 Kj/kg .K
There won't be any heat transmission because the process is isentropic.
So, Q = 0 KJ
Due to the isentropic nature of the operation, the ultimate entropy is determined by
s₂ = s₁ = 0.9803 Kj/kg .K
From the r-134a superheated refrigerant
at P₂ = 140 kpa & entropy s₂ = 0.9803 Kj/kg .K
the internal energy,u₂ = 227.77 K/kg
The work on the system is provided by
W = m x (u₂-u₁)
W = 0.7 x(227.77 - 263.87)
W = 0.7 x (-36.1)
W = -25.27 KJ
A negative number means the system is doing the task.
As a result, there won't be any heat transfer, and the system will only have used 25.27 KJ of energy.
Learn more about the work here:
https://brainly.com/question/20309171
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