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Question

# Temperature differences on the Rankine scale are identical to differences on the Fahrenheit scale, but absolute zero is given as $0 ^ { \circ } \mathrm { R }$. Find a second relationship converting temperatures $T_R$ of the Rankine scale to the temperatures $T_K$ of the Kelvin scale.

Solution

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The relation between $T_R$ and $T_F$ is:

\begin{align*} T_R&=T_F+459.67 \tag{Relation 1.}\\ \end{align*}

Relation for $T_F$ is:

\begin{align*} T_F&=T_R-459.67 \tag{Relation 2.}\\ \end{align*}

Use the following relation to convert temperature from degree Celsius to Kelvin scale:

\begin{align*} T_K&=T_C+273.15 \tag{Equation 1.}\\ \end{align*}

Expression for convert temperature from Fahrenheit to degree Celsius:

\begin{align*} T_C&=\frac{5}{9}-(T_F-32) \tag{Equation 2.}\\ \end{align*}

Now,

\begin{align*} T_C&=\frac{5}{9} (T_F-32) \tag{T_K=T_C+273.15.}\\ T_K&=\frac{5}{9}(T_F-32)+273.15 \tag{T_F=T_R-459.67.}\\ T_K&=\frac{5}{9} (T_R-459.67-32)+273.15\\ T_K&=\frac{5}{9} T_R\\ \end{align*}

So, the relation between $T_K$ and $T_R$ is given by:

$T_K=\frac{5}{9}T_R$

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