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Question
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.
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Solution
Given:-
Heat given to the system, ∆Q = 2625 cal
Increase in the internal energy of the system, ∆U = 5000 J
From the graph, we get
W = Area of the rectangle formed under line ab + Area under line bc
For line BC:-
Change in volume = 0
WBC = P Δ V = 0
∆W = Area of the rectangle
∆W = 200 × 103 × 0.03
= 6000 J
We know,
∆Q = ∆W + ∆U
⇒ 2625 cal = 6000 J + 5000 J
\[\Rightarrow J = \frac{11000}{2625} = 4 . 19 \text{ J/cal}\]
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