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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 0R0 ^ { \circ } \mathrm { R }. Find a second relationship converting temperatures TRT_R of the Rankine scale to the temperatures TKT_K of the Kelvin scale.

Solution

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The relation between TRT_R and TFT_F is:

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

Relation for TFT_F is:

TF=TR459.67\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:

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

Expression for convert temperature from Fahrenheit to degree Celsius:

TC=59(TF32)\begin{align*} T_C&=\frac{5}{9}-(T_F-32) \tag{Equation 2.}\\ \end{align*}

Now,

TC=59(TF32)TK=59(TF32)+273.15TK=59(TR459.6732)+273.15TK=59TR\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 TKT_K and TRT_R is given by:

TK=59TRT_K=\frac{5}{9}T_R

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