Solved Problem on Temperature
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One thermometer has the Celsius and Fahrenheit scales. Determine the temperature variation, on the Fahrenheit scale, which corresponds to a change of 20 °C.
Problem data:
- Variation of temperature on the Celsius scale: ΔtC = 20 °C.
The equation for convert temperature between Celsius and Fahrenheit scales is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\frac{t_{C}}{5}=\frac{t_{F}-32}{9}} \tag{I}
\end{gather}
\]
The initial and final temperatures on the Celsius scale are, respectively,
TC1 and TC2, the initial and final temperatures on
the Fahrenheit scale are, respectively, TF1 and
TF2, the expression (I) for these two states are
\[
\begin{gather}
\frac{t_{C1}}{5}=\frac{t_{F1}-32}{9} \tag{II}
\end{gather}
\]
\[
\begin{gather}
\frac{t_{C2}}{5}=\frac{t_{F2}-32}{9} \tag{III}
\end{gather}
\]
subtracting the expression (II) from (III), we have
\[
\begin{gather}
\frac{\left.
\begin{matrix}
\dfrac{t_{C2}}{5}=\dfrac{t_{F2}-32}{9}\\
\qquad\qquad\dfrac{t_{C1}}{5}=\dfrac{t_{F1}-32}{9} \qquad \text{(--)}
\end{matrix}
\right.}
{\dfrac{t_{C2}}{5}-\dfrac{t_{C1}}{5}=\dfrac{t_{F2}-32}{9}-\dfrac{t_{F1}-32}{9}}\\[6pt]
\frac{t_{C2}-t_{C1}}{5}=\frac{t_{F2}-32-\left(t_{F1}-32\right)}{9}\\[6pt]
\frac{t_{C2}-t_{C1}}{5}=\frac{t_{F2}-32-t_{F1}+32}{9}\\[6pt]
\frac{t_{C2}-t_{C1}}{5}=\frac{t_{F2}-t_{F1}}{9}
\end{gather}
\]
we have
\( \Delta t_{C}=t_{C2}-t_{C1} \)
and
\( \Delta t_{F}=t_{F2}-t_{F1} \)
the temperature variations on the scales Celsius and Fahrenheit
\[
\begin{gather}
\frac{\Delta t_{C}}{5}=\frac{\Delta t_{F}}{9}\\
\Delta t_{F}=\frac{9}{5}\Delta t_{C}
\end{gather}
\]
substituting the problem data, we have
\[
\begin{gather}
\Delta t_{F}=\frac{9}{5}\times 20\\
\Delta t_{F}=\frac{180}{5}
\end{gather}
\]
\[ \bbox[#FFCCCC,10px]
{\Delta t_{F}=36 °\;\text{F}}
\]
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .