Answer
The Range of Temperature on the surface of the Sun is $5500 ^{\circ}C to 6000^{\circ} C$.
Now, we have to convert $5500 ^{\circ}$C to Kelvins and $^{\circ} F$.
a) Converting to Kelvins:
K = $[ 5500^{\circ} C + 273.15 ^{\circ}] \times (1.0 K/1.0^{\circ}C)$ = 5773.15 K =$\approx$ 5773 K
b) Converting to $^{\circ} F$:
= [ $5500^{\circ}C \times (1.8^{\circ} F/1.0^{\circ}C)] + 32 ^{\circ} F$
= $9932^{\circ} F$
c) The famous astronomer in public lecture told that the temperature of the surface of sun is $10,000^{\circ}$. Here, she is referring to the Fahrenheit temperature Scale.
Work Step by Step
The Range of Temperature on the surface of the Sun is $5500 ^{\circ}C to 6000^{\circ} C$.
Now, we have to convert $5500 ^{\circ}$C to Kelvins and $^{\circ} F$.
The formulas for these conversions are as follows:
K = $x^{\circ}C + 273.15 ^{\circ}$ & $ ^{\circ} F = [ x^{\circ}C \times (1.8^{\circ} F/1.0^{\circ}C)] + 32 ^{\circ} F$
a) Converting to Kelvins:
K = $x^{\circ}C + 273.15 ^{\circ}$
K = $[ 5500^{\circ} C + 273.15 ^{\circ}] \times (1.0 K/1.0^{\circ}C)$ = 5773.15 K
b) Converting to $^{\circ} F:
?^{\circ} F$ = [$x^{\circ}C \times (1.8^{\circ} F/1.0^{\circ}C)] + 32 ^{\circ} F$
= [ $5500^{\circ}C \times (1.8^{\circ} F/1.0^{\circ}C)] + 32 ^{\circ} F
= 9900^{\circ} F + 32^{\circ} F$
= $9932^{\circ} F$
c) The famous astronomer in public lecture told that the temperature of the surface of sun is $10,000^{\circ}$. Here she is referring to the Fahrenheit temperature Scale.
