
A Centigrade and a Fahrenheit thermometer are dipped in boiling water. The water temperature is lowered until the Fahrenheit thermometer registers$140{}^\circ $. What is the fall in temperature as registered by the Centigrade thermometer?
A) $30{}^\circ $
B) $40{}^\circ $
C) $60{}^\circ $
D) $80{}^\circ $
Answer
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Hint:Temperature of a body, object, or solution is expressed as a number on a scale of temperature. Temperature is measured by two scales Celsius or Centigrade scale and Fahrenheit scale. we will first try to find out the relationship between the Centigrade and Fahrenheit scale with respect to ice point and steam points which have been given different numerical values in different scales of temperature.
Step by step solution:
Formula derived: In the Centigrade or Celsius system, the ice point is called $0{}^\circ C$, and the steam point $100{}^\circ C$, and their interval is divided into $100$equal parts or degrees.
On the Fahrenheit scale, the ice point and the steam point are called $32{}^\circ F$ and $212{}^\circ F$respectively and their intervals have been divided into $180$ degrees.
Thus, if temperature is $TC$ on the Celsius scale and $TF$ on the Fahrenheit scale, then
$\dfrac{TC}{100}=\dfrac{TF-32{}^\circ} {180}$
$\Rightarrow TC=100\dfrac{\left( TF-32{}^\circ \right)}{180}$
$\Rightarrow TC=\dfrac{5}{9}\left( TF-32{}^\circ \right)$
The above formula obtained is the relation between Celsius and Fahrenheit scale.
Given: Lowering in temperature on the Fahrenheit scale is $140{}^\circ F$ that is, $TF=140{}^\circ F$
Substituting this value in the formula above,
$TC=\dfrac{5}{9}\left( 140{}^\circ -32{}^\circ \right)$
$\Rightarrow TC=\dfrac{5}{9}\times 108{}^\circ \Rightarrow TC=5\times 12{}^\circ $
$\Rightarrow TC=60{}^\circ C$
We know the boiling point or steam point on the Celsius scale is$100{}^\circ C$. After lowering the temperature, the thermometer reads $60{}^\circ C$ on the Centigrade scale.
Hence, the fall in temperature in Centigrade will be, $100{}^\circ C-60{}^\circ C=40{}^\circ C$
The answer is $40{}^\circ C$
therefore option (B) is the correct option.
Note:Always try to solve the problem systematically starting very first from deriving the formula. Try to remember the formula to solve the problem quickly. Remember the fall in temperature is asked so we need to subtract the answer obtained in Celsius from$100{}^\circ C$ which is the boiling point or steam point of water on Centigrade scale.
Step by step solution:
Formula derived: In the Centigrade or Celsius system, the ice point is called $0{}^\circ C$, and the steam point $100{}^\circ C$, and their interval is divided into $100$equal parts or degrees.
On the Fahrenheit scale, the ice point and the steam point are called $32{}^\circ F$ and $212{}^\circ F$respectively and their intervals have been divided into $180$ degrees.
Thus, if temperature is $TC$ on the Celsius scale and $TF$ on the Fahrenheit scale, then
$\dfrac{TC}{100}=\dfrac{TF-32{}^\circ} {180}$
$\Rightarrow TC=100\dfrac{\left( TF-32{}^\circ \right)}{180}$
$\Rightarrow TC=\dfrac{5}{9}\left( TF-32{}^\circ \right)$
The above formula obtained is the relation between Celsius and Fahrenheit scale.
Given: Lowering in temperature on the Fahrenheit scale is $140{}^\circ F$ that is, $TF=140{}^\circ F$
Substituting this value in the formula above,
$TC=\dfrac{5}{9}\left( 140{}^\circ -32{}^\circ \right)$
$\Rightarrow TC=\dfrac{5}{9}\times 108{}^\circ \Rightarrow TC=5\times 12{}^\circ $
$\Rightarrow TC=60{}^\circ C$
We know the boiling point or steam point on the Celsius scale is$100{}^\circ C$. After lowering the temperature, the thermometer reads $60{}^\circ C$ on the Centigrade scale.
Hence, the fall in temperature in Centigrade will be, $100{}^\circ C-60{}^\circ C=40{}^\circ C$
The answer is $40{}^\circ C$
therefore option (B) is the correct option.
Note:Always try to solve the problem systematically starting very first from deriving the formula. Try to remember the formula to solve the problem quickly. Remember the fall in temperature is asked so we need to subtract the answer obtained in Celsius from$100{}^\circ C$ which is the boiling point or steam point of water on Centigrade scale.
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