JEE Mains · Physics · STD 11 - 11. thermodynamics
A gas can be taken from \(A\) to \(B\) via two different processes \(ACB\) and \(ADB\). When path \(ACB\) is used \(60\, J\) of heat flows into the system and \(30\, J\) of work is done by the system. If path \(ADB\) is used work down by the system is \(10\, J\). the heat flow into the system in path \(ADB\) is ..... \(J\)
- A \(40\)
- B \(80\)
- C \(100\)
- D \(20\)
Answer & Solution
Correct Answer
(A) \(40\)
Step-by-step Solution
Detailed explanation
As temperature at point \(A\) and \(C\) is same. \(\therefore\) Internal energy change will be same. \(\mathrm{Q}-\mathrm{W}=\mathrm{Q}^{\prime}-\mathrm{W}^{\prime}\) \(60-30=Q^{\prime}-10\) \(Q^{\prime}=40 \mathrm{J}\)
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