A system is initially at conditions of

P= 200 kPa and V= 0.1 m^3

Heat is now added to the system, and the system expands in a constant pressure process. At the end of the process the volume has doubled.

(a) Calculate the work transferred during the process.

(b) During the process the internal energy of the system increases by 200 kJ; kinetic and potential energy remains constant. Determine the net heat transfer during the process.

Question

contestada

Respuesta :

Manetho Manetho · Answer 1 · 2019-07-24T11:43:42+02:00

Answer:

a)W=20 KJ

b) ΔQ= 220 KJ

Explanation:

Given:

V₁=0.1 m^3, P₁=200 kPa and heat is added to the system such that system expands under constant pressure.

Therefore V₂= 2V₁= 0.2 m^3

a) Work transfer W= P(V₂-V₁)= [tex]200\times(0.2-0.1)\times10^{5} = 2\times10^4 joules[/tex]

W=20 KJ

b) internal energy change ΔU= 200 KJ

from first law we know that ΔQ(net heat transfer)= ΔU + W

ΔQ= [tex]200\times10^3 +2\times10^4[/tex]

ΔQ=[tex]22\times10^4 J[/tex]

ΔQ= 220 KJ