SOLUTION:• For Celsius[tex]\begin{aligned} ^{\circ}\text{C} & = \frac{5}{9}(^{\circ}\text{F} - 32) \\ & = \frac{5}{9}(83 - 32) \\ & = \frac{5}{9}(51) \\ & = \frac{255}{9} \\ & = \boxed{28.33^{\circ}\text{C}} \end{aligned}[/tex]• For Kelvin[tex]\begin{aligned} \text{K} & = (^{\circ}\text{C}) + 273.15 \\ & = 28.33 + 273.15 \\ & = \boxed{\text{301.48 K}} \end{aligned}[/tex]