Question

A faulty thermometer reads $5^\circ C$ in melting ice and $99^\circ C$ in dry steam. Find the correct temperature in Fahrenheit scale if this faulty thermometer reads $52^\circ C$.
$84^\circ F$
$122^\circ F$
$56^\circ F$
$88^\circ F$

Answer 1

Use the conversion formula to convert the scales of measurement of the temperature. The lowest temperature at which the ice melts in Celsius scale is $0^\circ C$ and highest temperature of the steam is $100^\circ C$. Use the conversion formula to convert the temperature from Celsius to Fahrenheit.

Formula used:
$\dfrac{{{T_s} - {{\left( {MP} \right)}_s}}}{{U{T_s} - L{T_s}}} = \dfrac{{{T_C} - {{\left( {MP} \right)}_C}}}{{U{T_C} - L{T_C}}}$

Here, ${\left( {MP} \right)_s}$ is the melting point of ice of ${T_s}$ scale, ${\left( {MP} \right)_C}$ is the melting point of ice of Celsius scale, $U{T_s}$ is the upper temperature and $L{T_s}$ is the lowest temperature of the scale ${T_s}$, and $U{T_C}$ is the upper temperature and $L{T_C}$ is the lowest temperature of the scale ${T_C}$.

Complete step by step answer: Let the temperature scale of this faulty thermometer is ${T_s}$ and the temperature scale of Celsius scale is ${T_C}$.

Therefore, we can write from the conversion formula to convert the two scales as follows,
$\dfrac{{{T_s} - {{\left( {MP} \right)}_s}}}{{U{T_s} - L{T_s}}} = \dfrac{{{T_C} - {{\left( {MP} \right)}_C}}}{{U{T_C} - L{T_C}}}$ …… (1)

Here, ${\left( {MP} \right)_s}$ is the melting point of ice of ${T_s}$ scale, ${\left( {MP} \right)_C}$ is the melting point of ice of Celsius scale, $U{T_s}$ is the upper temperature and $L{T_s}$ is the lowest temperature of the scale ${T_s}$, and $U{T_C}$ is the upper temperature and $L{T_C}$ is the lowest temperature of the scale ${T_C}$.

Substitute $52^\circ C$ for ${T_s}$, $5^\circ C$ for ${\left( {MP} \right)_s}$, $99^\circ C$ for $U{T_s}$, $5^\circ C$ for $U{T_s}$, $0^\circ C$ for ${\left( {MP} \right)_C}$, $100^\circ C$ for $U{T_C}$ and $0^\circ C$ for $L{T_C}$ in the above equation.
$\dfrac{{52 - 5}}{{99 - 5}} = \dfrac{{{T_C} - 0}}{{100 - 0}}$
$\Rightarrow {T_C} = 100 \times \dfrac{{47}}{{94}}$
$\therefore {T_C} = 50^\circ C$

Convert the Celsius temperature into Fahrenheit temperature using the below formula,
${T_F} = \dfrac{9}{5}{T_C} + 32$

Substitute ${T_C} = 50^\circ C$ in the above equation.
${T_F} = \dfrac{9}{5}\left( {50} \right) + 32$
$\Rightarrow {T_F} = 90 + 32$

So, the correct answer is option (B).

Note: You can also convert the Celsius scale to Fahrenheit scale using equation (1) by substituting upper and lower limits of the temperatures as $32^\circ F$ and $212^\circ F$ respectively.