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.

When a system is taken from state B to state A along path BDA as shown

1.	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.
Answer» `q_(AC)=-20J` & `q_(BC)=-50J` `q_(AC)=-20 & q_(BC)=50J` `q_(AC)=20J&q_(DB)=50J` `q_(AC)=20J & q_(BC)=-50J` Solution :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`