The heat absorbed by a system in going throug | vedclass

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(A) For a cyclic process, the change in internal energy is $\Delta U = 0$.According to the first law of thermodynamics, $\Delta Q = \Delta U + W$. Sin

Vedclass · Accepted Answer

$61.6$

Vedclass · Answer

(A) For a cyclic process, the change in internal energy is $\Delta U = 0$.According to the first law of thermodynamics, $\Delta Q = \Delta U + W$. Since $\Delta U = 0$, the heat absorbed $\Delta Q$ is equal to the work done $W$, which is the area enclosed by the $P-V$ curve.The $P-V$ curve is a circle with diameter along the pressure axis $d_P = (340 - 60) \,kPa = 280 \,kPa = 280 	imes 10^3 \,Pa$ and diameter along the volume axis $d_V = (340 - 60) \,cc = 280 \,cm^3 = 280 	imes 10^{-6} \,m^3$.The radii are $r_P = 140 	imes 10^3 \,Pa$ and $r_V = 140 	imes 10^{-6} \,m^3$.The area of the circle is $A = \pi 	imes r_P 	imes r_V$.$W = \pi 	imes (140 	imes 10^3 \,Pa) 	imes (140 	imes 10^{-6} \,m^3) = \pi 	imes 140 	imes 140 	imes 10^{-3} \,J = \pi 	imes 19.6 \,J \approx 3.14159 	imes 19.6 \,J \approx 61.575 \,J \approx 61.6 \,J$.

The heat absorbed by a system in going through the given cyclic process is: (in $\,J$)

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