Answer:
Q= 12 KJ
Heat enters to the system.
Explanation:
Given that
m= 5 kg
P₁ = 400 KPa
T₁ = 30⁰C = 303 K
Work done by the system ,W₁= 15 KJ
Work is done on the system W₂ = - 3 KJ'
Given that expansion is isothermal ,that is why temperature of the gas will remain constant.We know that internal energy of the ideal gas only depends on the temperature of the gas.That is why change in the internal energy of the gas will be zero because change in the temperature is zero.
From first law of thermodynamics
Q = W + ΔU
Q= =Heat ,W= Net Work ,ΔU =Change in the internal energy
Here ΔU = 0
Q= W
Q= 15 - 3 KJ
Q= 12 KJ
Heat enters to the system.
Note -
Work done on the system will be taken as negative and work done by the system taken as positive.
Heat added to the system will be taken as positive heat leaving from the system taken as negative.
