Sum

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

W_{BC} = P Δ V = 0

∆W = Area of the rectangle

∆W = 200 × 10^{3} × 0.03

= 6000 J

We know,

∆Q = ∆W + ∆U

⇒ 2625 cal = 6000 J + 5000 J

\[\Rightarrow J = \frac{11000}{2625} = 4 . 19 \text{ J/cal}\]

Concept: Heat, Internal Energy and Work

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