Question

What is the temperature for which the readings on Kelvin and Fahrenheit scales are the same?

Answer 1

Hint:For solving this question we should be well aware about the relationship between kelvin and Fahrenheit. We know that to convert Kelvin into Fahrenheit mathematically we can use below formula.
${\text{F = }}\left( {{\text{K - 273}}} \right)\, \times \,\dfrac{9}{5}\, + \,32$

Step by step Solution:
For finding the point of coincidence of these two scales, we need to first understand the conversions. In any scale (that we define or is pre-defined),
[Temperature - Freezing point of water (in this scale)]/ [Boiling point of water (in this scale) - Freezing point of water (in this scale)] is a constant.
So, mathematically it can be expressed as
$\dfrac{{\left( {{\text{K - 273}}{\text{.15}}} \right)}}{{\left( {373.15 - 273.15} \right)}}\, = \,\dfrac{{\left( {{\text{C}} - 0} \right)}}{{\left( {100 - 0} \right)}}\, = \,\dfrac{{\left( {{\text{F - 32}}} \right)}}{{\left( {212 - 32} \right)}}\, = $ constant
Where “K” is the temperature in kelvin, C is the temperature in Celsius and F is the temperature in Fahrenheit.
This in turn gives us the relationship among three scales (Kelvin, Celsius and Fahrenheit),
$\dfrac{{\left( {{\text{K - 273}}{\text{.15}}} \right)}}{{100}}\, = \,\dfrac{{\text{C}}}{{100}}\, = \,\dfrac{{\left( {{\text{F - 32}}} \right)}}{{180}}$
For the scope of this question we just need the relationship between Kelvin and Fahrenheit so we will use
$\dfrac{{\left( {{\text{K - 273}}{\text{.15}}} \right)}}{{100}}\, = \,\,\dfrac{{\left( {{\text{F - 32}}} \right)}}{{180}}$
Assuming at temperature (T) both kelvin and Fahrenheit temperature becomes equal so replacing “K” and “F” both with “T”
$\dfrac{{\left( {{\text{T - 273}}{\text{.15}}} \right)}}{{100}}\, = \,\,\dfrac{{\left( {{\text{T - 32}}} \right)}}{{180}}$
$\dfrac{{\left( {{\text{T - 273}}{\text{.15}}} \right)}}{5}\, = \,\,\dfrac{{\left( {{\text{T - 32}}} \right)}}{9}$
Cross multiplying the equations we get,
$9{\text{T - 9 }} \times {\text{ 273}}{\text{.15 = 5T - 160}}$
Placing variables on one side and solving further
\[4{\text{T = 2458}}{\text{.35 - 160}}\]
So, \[{\text{T = }}\dfrac{{2298.35}}{4}\]
\[{\text{T = }}574.587\]
Hence, we can write \[574.587\] Kelvin is equal to \[574.587\] Fahrenheit.

Note:Kelvin scale got its name from British inventor and scientist William Thomson who was also called as Lord Kelvin. Kelvin temperature scale is an absolute temperature scale with zero at absolute zero. Fahrenheit scale was introduced and named after Daniel Gabriel Fahrenheit.