Question

i) An object has a temperature of ${110^ \circ }F$. What is the temperature in degrees Celsius?
ii) An object has a temperature of ${100^ \circ }F$. What is the temperature in kelvin?
iii) The temperature of two bodies differ by ${1^ \circ }C$ . How much do they differ on the Kelvin scale?

Answer 1

Hint: For solving this question we’ll have to recall the conversions used for converting it from one scale to another. Consider a temperature T which has the same value in all the scales. The conversion from Fahrenheit to Celsius can be given as: ${T^ \circ }C = ({T^ \circ }F - 32)\dfrac{5}{9}$
To convert Fahrenheit to Kelvin the relation can be given as: $TK = ({T^ \circ }F - 32)\dfrac{5}{9} + 273.15$

Complete answer: We have been an object at a certain temperature in degree Fahrenheit.
i) The temperature of the given object is ${110^ \circ }F$ . To convert it into degree Celsius we’ll use the formula ${T^ \circ }C = ({T^ \circ }F - 32)\dfrac{5}{9}$
Substituting the value of ${T^ \circ }F = {110^ \circ }F$
${T^ \circ }C = (110 - 32) \times \dfrac{5}{9} = 78 \times \dfrac{5}{9} = {43.33^ \circ }C$
Hence ${110^ \circ }F$ in Celsius scale is ${43.33^ \circ }C$
ii) The temperature of the object is given as ${100^ \circ }F$ . To convert it into Kelvin we’ll use the formula $TK = ({T^ \circ }F - 32)\dfrac{5}{9} + 273.15$
Substituting the value of ${T^ \circ }F = {100^ \circ }F$
$TK = (100 - 32) \times \dfrac{5}{9} + 273.15 = 68 \times \dfrac{5}{9} + 273.15$
$TK = 37.77 + 273.15 = 310.92K$
Hence ${110^ \circ }F$ in Celsius scale is 310.29K.
iii) Given that the two bodies differ by ${1^ \circ }C$. Consider the temperature of A to be ${0^ \circ }C$ and that of B to be ${1^ \circ }C$.
The conversion factor for Celsius to kelvin is: ${T^ \circ }C = (273 + T)K$
Hence for ${0^ \circ }C$ to kelvin can be given as: ${0^ \circ }C = 273 + 0 = 273K$ --(1)
For ${1^ \circ }C$ kelvin can be given as: ${1^ \circ }C = 273 + 1 = 274K$
The difference between the two temperatures in Kelvin will be: $274 - 273 = 1K$

Note:
The difference in degree Celsius will be equal to the difference in Kelvin. If the difference in Degree Celsius is ${5^ \circ }$ then the difference in kelvin will also be 5 K only. Hence, we can say that $\Delta K = {\Delta ^ \circ }C$.
Though we commonly use degrees Celsius and Fahrenheit to measure the temperatures, the SI unit of temperature is Kelvin. The lowest known temperature is $ - {273.15^ \circ }C = 0K$ which is known as the absolute zero temperature.