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.
Figure

Answer 1

The value of J is 4.19 j/cal.

Explanation:

Step 1:

∆Q = 2625 cal

∆U = 5000 J

∆Q  is Heat given to the system,

∆U   is Increase in the internal energy of the system,  

Step 2:

W = A rectangle area formed under ab line  + Area under BC line  

For BC line:

ΔV= 0

ΔV is change in volume  

WBC=PΔV=0

∆W = rectangle Area

$\begin{aligned}\Delta W &=200 \times 10^{3} \times 0.03 \\&=6000 \mathrm{J}\end{aligned}$

Step 3:

We know that

$\begin{array}{l}{\Delta Q=\Delta W+\Delta U} \\{2625 c a l=6000 J+5000 J}\end{array}$

$J=\frac{11000}{2003}=4.19 \mathrm{J} / \mathrm{cal}$

Answer 2

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Please refer the above attachment