Gas with in a chamber, passes through the cycle shown in (figure). Determine the net heat added to the system during process AB is `Q_(AB)= 20J`. No heat is transferred during process BC and net work done during the cycle is `15J`

Explanation

According to first law of thermodynamics.
In process CA,
`dQ_(CA)=dU_(CA)+dW_(CA)`
In process AB
`dQ_(AB)=dU_(AB)+dW_(AB)`, and
In process BC
`dQ_(BC)=dU_(BC)+dW_(BC)`
Adding the three equations, we get
`dQ_(CA)+dQ_(AB)+dQ_(BC)= (dU_(CA)+dU_(AB)+dU_(BC))`
`+(dW_(CA)+dW_(AB)+dW_(BC)`....(i)
As the process is cyclic, `dU=dU_(CA)+dU_(AB)+dU_(BC)=0`
From (i) `dQ_(CA)+20+0=0+15`
`dQ_(CA)=15-20= -5J`
This is the net heat added to the system.