Question

When a system is taken from state `B` to state `A` along path `BDA` as shown in figure below, `60 J` of heat flows out of the system and `10J` of work is doen on path `ACB` , then the heat corresponding to the processes `AC` and `BC` is respectively.

A. `q_(AC)=-20J` & `q_(BC)=-50J`
B. `q_(AC)=-20 & q_(BC)=50J`
C. `q_(AC)=20J&q_(DB)=50J`
D. `q_(AC)=20J & q_(BC)=-50J`

Answer 1

Correct Answer - 4
ln ACB process, AC process is isochoric, so `W_(AC)=0`.
So, `" "DeltaE_(BC)=q_(ac)+W_(ac)`
`-30=q_(BC)+(20)`
`q_(BC)=-50J`
Now, `" "q_(AB)+q_(AC)+q_(CB)`
`70=q_(AC)+50(` since for path `BDA,E_(A)-E_(B)+q+W=-60+10=-50`
J and `W_(ACB)=-20J,` So, `q_(AB)=DeltaE_(AB)-W_(ACB)=50+20=70J)`
`q_(AC)=20J`