Answer
(a) $27.3s$
(b) $61.5s$
(c) $246s$
(d) the water is boiling
Work Step by Step
(a) The time taken for the system to reach point B from point A is given as
$t_{AB}=\frac{Q_A+mL_f-Q_A}{\frac{\Delta Q}{\Delta T}}$
We plug in the known values to obtain:
$t_{AB}=\frac{(1Kg)(33.5\times 10^4J/Kg)}{12250J/s}$
$\implies t_{AB}=27.3s$
(b) We know that the time taken from B to C is given as
$t_{BC}=\frac{mC_{water}\Delta T}{\frac{\Delta Q}{\Delta T}}$
We plug in the known values to obtain:
$t_{BC}=\frac{(1Kg)(4186J/Kg.K)(100C^{\circ}-0^C{\circ})}{12,250J/s}$
$\implies t_{BC}=34.2s$
Now time taken from point A to point C is
$t_{AC}=t_{AB}+t_{BC}$
$\implies t_{AC}=27.3s+34.2s=61.5s$
(c) The time taken by the system to reach from point C to point D is
$t_{CD}=\frac{mL_V}{\frac{\Delta Q}{\Delta T}}$
We plug in the known values to obtain:
$t_{CD}=\frac{(1Kg)(22.6\times 10^5J/Kg)}{12,250J/s}=184.5s$
Now, the time taken to reach point D from point A is
$t_{AD}=t_{AB}+t_{BC}+t_{CD}=27.3s+34.2s+184.5s=246s$
(d) We know that at $t=63seconds$, the system is in the CD portion and we observe that the water should be boiling at this time.
