Solved Problem on Temperature

The difference between the values indicated in a thermometer in the Fahrenheit scale and another in the Celsius scale are equal to 40 for the same temperature. What are the temperatures read in the two thermometers?

Problem data:

Difference between the thermometer values: Δt = 40.

Solution

The problem gives the difference between temperatures on the Fahrenheit and Celsius scales is equal to

\[ \begin{gather} \Delta t=t_{F}-t_{C}=40 \tag{I} \end{gather} \]

The equation to convert temperature in Celsius into Fahrenheit is

\[ \bbox[#99CCFF,10px] {\frac{t_{C}}{5}=\frac{t_{F}-32}{9}} \]

writing how

\[ \begin{gather} t_{C}=\frac{5}{9}(t_{F}-32) \tag{II} \end{gather} \]

substituting the expression (II) into the condition (I), we have

\[ t_{F}-\frac{5}{9}(t_{F}-32)=40 \]

multiplying the expression by 9, we obtain

\[ \begin{gather} \qquad\qquad t_{F}-\frac{5}{9}(t_{F}-32)=40 \qquad (\times 9)\\ 9t_{F}-\cancel{9}\times \frac{5}{\cancel{9}}\times (t_{F}-32)=9\times40\\ 9t_{F}-5\times (t_{F}-32)=360\\ 9t_{F}-5t_{F}+5\times 32=360\\ 4t_{F}+160=360\\ 4t_{F}=360-160\\ 4t_{F}=200\\ t_{F}=\frac{200}{4} \end{gather} \]

\[ \bbox[#FFCCCC,10px] {t_{F}=50\;\text{°F}} \]

substituting this value in the expression (II), we have

\[ \begin{gather} t_{C}=\frac{5}{9}\times (50-32)\\ t_{C}=\frac{5}{\cancel{9}}\times \cancelto{2}{18}\\ t_{C}=5\times 2 \end{gather} \]

\[ \bbox[#FFCCCC,10px] {t_{C}=10\;\text{°C}} \]

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