A steel alloy is rapidly cooled to $\pu{600^oC}$, held for $\pu{4 sec}$, rapidly cooled to $\pu{450^oC}$, held for $\pu{10 sec}$ and quenched to room temperature. What phases are produced?

Why is the time reset back to $0$ when you cool to $\pu{450^oC}$? Is it because I am only considering the remaining austenite?

the graph in question


The given phase diagram provides some information on the time-temperature relationship for the formation of pearlite (P), bainite (B), and martensite (M) from austenite (A) form, which is known as known as gamma-phase iron ($\gamma$-$\ce{Fe}$). For example, It would take $\pu{10 sec}$ to 100% austenite to the 50% austenite-to-pearlite transformation to complete.

As following diagram indicate, rapid cooling of eutectoid composition of steel alloy (100% A) to $\pu{600 ^\circ C}$ stays as 100% A. If the temperature held at $\pu{600 ^\circ C}$ for $\pu{4 sec}$, austenite-to-pearlite transformation occur to ~25% (see the point noted at the diagram):

Iorn-Iorn carbide phase diagram

During the second rapid cooling from $\pu{600 ^\circ C}$ to $\pu{450 ^\circ C}$ (time $\approx 0$), there is no additional transformation (see the second point noted at the diagram with red line). At $\pu{450 ^\circ C}$, the time is start to count again at zero second, so that the time is moved to $t=0$ point (this is also because isotherm conversion will be started with unconverted austenite, which is now 75% of original specimen). While holding at $\pu{450 ^\circ C}$ for $\pu{10 sec}$, approximately 50% of this remaining unreacted 75% A (or ~37.5% of the original specimen) will transform to bainite (see the point noted in the diagram). Finally, upon cooling to room temperature, the remaining 37.5% A (of the original specimen) transforms to martensite. Thus, the final micro-structure of given steel alloy consists of about 25% pearlite, 37.5% bainite, and 37.5% martensite (no austenite).

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