Consider a cycle followed by an engine (figure.)
1 to 2 is isothermal
2 to 3 is adiabatic
3 to 1 is adiabatic
Such a process does not exist, because
Consider a heat engine as shown in figure. $Q_1$ and $Q_2$ are heat added both to $T_1$ and heat taken from $T_2$ in one cycle of engine. $W$ is the mechanical work done on the engine.
If W > 0, then possibilities are
Can a system be heated and its temperature remains constant?
Yes, this is possible when the entire heat supplied to the system is utilised in expansion. i.e., its working against the surroundings.
A system goes from $P$ to $Q$ by two different paths in the $p-V$ diagram as shown in figure. Heat given to the system in path 1 is 1000 J . The work done by the system along path 1 is more than path 2 by 100 J . What is the heat exchanged by the system in path 2?
$$ \begin{aligned} \text { For path 1, } \quad \text { Heat given } Q_1 & =+1000 \mathrm{~J} \\ \text { Work done } & =W_1 \text { (let) } \end{aligned}$$
For path 2,
Work done $$\left(W_2\right)=\left(W_1-100\right) \mathrm{J}$$
Heat given $Q_2=$ ?
$$\begin{aligned} &\text { As change in internal energy between two states for different path is same. }\\ &\begin{aligned} \therefore \quad & \Delta U =Q_1-W_1=Q_2-W_2 \\ \Rightarrow \quad & 1000-W_1 =Q_2-\left(W_1-100\right) \\ \Rightarrow \quad & Q_2 =1000-100=900 \mathrm{~J} \end{aligned} \end{aligned}$$
If a refrigerator's door is kept open, will the room become cool or hot? Explain.
If a refrigerator's door is kept open, then room will become hot, because amount of heat removed would be less than the amount of heat released in the room.