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

Hint: 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.