Answer
For Al, $660.4^{\circ}C$ and $1221^{\circ}F$
For Ag, $961.9^{\circ}C$ and $1763^{\circ}F$
Work Step by Step
For Al,
$933.6\, K=(933.6-273.15) ^{\circ}C=660.4^{\circ}C$
$=[(660.4\times\frac{9}{5})+32]^{\circ}F$
$=1221^{\circ}F$
For Ag,
$1235.1\, K=(1235.1-273.15)^{\circ}C=961.9^{\circ}C$
$=[(961.9\times\frac{9}{5}) +32]^{\circ}F$
$=1763^{\circ}F$
