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Q. 2.4

Define a new temperature scale, say °F, in which the steam point and ice point are 212°F and 32°F respectively. The °F reading on this scale is a number of degrees on a corresponding absolute temperature scale (also called Rankine scale). What is this temperature at 10°F ?

Verified Solution

\begin{aligned}\frac{T_s}{T_i} &=1.366 \\ \\T_s-T_i &=212-32=180 \\ \\(1.366-1) T_i &=180\end{aligned}

\begin{aligned}&T_i=\frac{180}{0.366}=491.7^{\circ} F \text { abs or } 491.7^{\circ} R \\ \\&T_s=180+491.7=671.7^{\circ} R\end{aligned}

T°R = (T°F – 32) + 491.7 = t°F + 459.7

At 10°F, T°R = 10 + 459.7 = 469.7