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(A) For a cyclic process,the change in internal energy $\Delta U = 0$. According to the first law of thermodynamics,$\Delta Q = \Delta W$.$1$. For pro

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$-40$

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(A) For a cyclic process,the change in internal energy $\Delta U = 0$. According to the first law of thermodynamics,$\Delta Q = \Delta W$.$1$. For process $1 \rightarrow 2$ (isothermal): $\Delta U_{1 \rightarrow 2} = 0$,so $\Delta Q_{1 \rightarrow 2} = \Delta W_{1 \rightarrow 2}$.$2$. For process $2 \rightarrow 3$ (isochoric): $\Delta W_{2 \rightarrow 3} = 0$. Given that $40 \ J$ of heat leaves the system,$\Delta Q_{2 \rightarrow 3} = -40 \ J$. From the first law,$\Delta U_{2 \rightarrow 3} = \Delta Q_{2 \rightarrow 3} - \Delta W_{2 \rightarrow 3} = -40 - 0 = -40 \ J$.$3$. For process $3 \rightarrow 1$ (adiabatic): $\Delta Q_{3 \rightarrow 1} = 0$. From the first law,$\Delta U_{3 \rightarrow 1} = \Delta Q_{3 \rightarrow 1} - \Delta W_{3 \rightarrow 1} = -\Delta W_{3 \rightarrow 1}$.Since the total change in internal energy for the cycle is zero: $\Delta U_{1 \rightarrow 2} + \Delta U_{2 \rightarrow 3} + \Delta U_{3 \rightarrow 1} = 0$.$0 + (-40) + (-\Delta W_{3 \rightarrow 1}) = 0$.Therefore,$\Delta W_{3 \rightarrow 1} = -40 \ J$.

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

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