Chemistry (7th Edition)

$1474$ $^{\circ}F$ is equal to $801$ $^{\circ}C$ and $1074$ $K$.
We find $^{\circ}C = (\frac{5^{\circ}C }{9^{\circ}F})(1474^{\circ}F-32^{\circ}F) = 801$ $^{\circ}C$ $K = ^{\circ}C + 273.15 = 801 + 273.15 = 1074$ $K$