In the process $2 \rightarrow 3, \Delta W_{2 \rightarrow 3}=0$
In the process $3 \rightarrow 1, \Delta Q_{3 \rightarrow 1}=0$
The complete process being cyclic, $\Delta U=0$
Hence $\Delta Q=\Delta W \Rightarrow \Delta Q_{1 \rightarrow 2}+\Delta Q_{2 \rightarrow 3}+\Delta Q_{3 \rightarrow 1}=\Delta W_{1 \rightarrow 2}+\Delta W_{2 \rightarrow 3}+\Delta W_{3 \rightarrow 1} \Rightarrow$
$\Delta Q_{2 \rightarrow 3}=\Delta W_{3 \rightarrow 1}=-40 J$
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[Given: The acceleration due to gravity $g=10 \mathrm{~ms}^{-2}$ and $\pi=3.14$ ]
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($1$) Considering the air flow to be streamline, the steady mass flow rate of air exiting the chimney is
. . . . .$\mathrm{gm} \mathrm{s}^{-1}$.
($2$) When the chimney is closed using a cap at the top, a pressure difference $\Delta P$ develops between the top and the bottom surfaces of the cap. If the changes in the temperature and density of the hot air, due to the stoppage of air flow, are negligible then the value of $\Delta P$ is. . . . .$\mathrm{Nm}^{-2}$.
