When a system is taken from state `A` to state `B` along path `ACB` as shown in figure below, `80 J` of heat flows into the system and the system does `30 J` of work. If `E_(D)-E_(A)= -40 J`, the heat absorbed in the processes `AD` and `DB` are respectivelyA. `q_(AD)=30J` and `q_(DB)= - 90 J`B. `q_(AD)=-60J` and `q_(DB)= 30 J`C. `q_(AD)=30J` and `q_(DB)= 90 J`D. `q_(AD)=-30J` and `q_(DB)= 90 J`

When a system is taken from state `A` to state `B` along path `ACB` as

1.	When a system is taken from state `A` to state `B` along path `ACB` as shown in figure below, `80 J` of heat flows into the system and the system does `30 J` of work. If `E_(D)-E_(A)= -40 J`, the heat absorbed in the processes `AD` and `DB` are respectivelyA. `q_(AD)=30J` and `q_(DB)= - 90 J`B. `q_(AD)=-60J` and `q_(DB)= 30 J`C. `q_(AD)=30J` and `q_(DB)= 90 J`D. `q_(AD)=-30J` and `q_(DB)= 90 J`
Answer» Correct Answer - D In `ADB` process, `DB` process is isochoric so `w_(DB)=0` So, `Delta E_(AD)=q_(AD)+w_(AD)` `-40=q_(AD)+(-10), q_(AD)= -30 J` Now, `q_(AB)=q_(AD)+q_(DB), 60 =-30+ q_(DB)` `q_(DB)=90 J`

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