A student records Delta Q, Delta U and Delta | vedclass

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Question

(D) For a thermodynamic cycle,the change in internal energy $\Delta U_{total} = 0$. Thus,$\Delta U_{AB} + \Delta U_{BC} + \Delta U_{CA} = 0$.Using the

Vedclass · Accepted Answer

$\Delta U_{CA} = -160 \ J$

Vedclass · Answer

(D) For a thermodynamic cycle,the change in internal energy $\Delta U_{total} = 0$. Thus,$\Delta U_{AB} + \Delta U_{BC} + \Delta U_{CA} = 0$.Using the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$,so $\Delta U = \Delta Q - \Delta W$.For process $AB$: $\Delta U_{AB} = \Delta Q_{AB} - \Delta W_{AB} = 150 \ J - 40 \ J = 110 \ J$.For process $BC$: $\Delta W_{BC} = \Delta Q_{BC} - \Delta U_{BC} = 10 \ J - 50 \ J = -40 \ J$.Now,$\Delta U_{CA} = -(\Delta U_{AB} + \Delta U_{BC}) = -(110 \ J + 50 \ J) = -160 \ J$.Thus,option $D$ is correct.

Process	$AB$	$BC$	$CA$
$\Delta W$	$40 \ J$	$ $	$30 \ J$
$\Delta U$	$ $	$50 \ J$	$ $
$\Delta Q$	$150 \ J$	$10 \ J$	$ $

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$A$ student records $\Delta Q, \Delta U$ and $\Delta W$ for a thermodynamic cycle $A \to B \to C \to A$. Certain entries are missing. Find the correct entry from the following options.Process$AB$$BC$$CA$$\Delta W$$40 \ J$$ $$30 \ J$$\Delta U$$ $$50 \ J$$ $$\Delta Q$$150 \ J$$10 \ J$$ $