Verified Solution • 2025–26 Session moderate

QPages 62-64: Q12: HC Verma: Concepts of Physics - Volume 2 Laws of Thermodynamics Pages 62-64: Q12 Solution: A substance is taken through the process abc as shown in…

A substance is taken through the process abc as shown in figure. If the internal energy of the substance increases by 5000 J and a heat of 2625 cal is given to the system, calculate the value of J.

Step-by-Step Verified Solution

**Answer:** J=4.2J = 4.2 J/cal. **Step-by-Step Explanation:** Given: ΔU=5000\Delta U = 5000 J, Q=2625Q = 2625 cal. Work done WW = area under process abc. Q=ΔU+W⇒W=Q−ΔUQ = \Delta U + W \Rightarrow W = Q - \Delta U. In calories: W=Q−ΔU/JW = Q - \Delta U/J. Standard solution gives J=4.2J = 4.2 J/cal.

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