Answer
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Hint:In this question, the concept of the unit conversion factor is used. First, we consider the expression for the body temperature to convert Fahrenheit into degree then substitute the value of Fahrenheit to calculate the value of temperature in degree Celsius.
Complete step by step solution:
In this question, the body temperature of a patient is \[104^\circ {\text{F}}\]. We need to calculate the body temperature in Celsius scale.
As we know normally available thermometers in the market are calibrated in either Fahrenheit $\left( {^\circ {\text{F}}} \right)$ or degree Celsius $\left( {^\circ {\text{C}}} \right)$ depending on the country, where it is being sold. The United States follows Fahrenheit measurement whereas most other countries follow degree Celsius.
We now know that an increase in temperature corresponds to an increase in the average kinetic energy of atoms and molecules. A result of that increased motion is that the average distance between atoms and molecules increases as the temperature increases. This phenomenon, known as thermal expansion, is the basis for temperature measurement by a liquid thermometer.
Let us assume, the temperature of the scale Fahrenheit $\left( {^\circ {\text{F}}} \right)$ or degree Celsius $\left( {^\circ {\text{C}}} \right)$.
We know that
$9C = 5\left( {F - 32} \right)......\left( 1 \right)$
Where, $C$ is temperature in Celsius and $F$ is in Fahrenheit.
As we know that the temperature that temperature is Fahrenheit is\[104^\circ {\text{F}}\]
We are putting the value of Fahrenheit that is \[104^\circ {\text{F}}\] in equation (1).
$ \Rightarrow 9C = 5\left( {104 - 32} \right)$
Now we subtract the value of Fahrenheit $104$ from $32$ then multiply it by $5$ and simplify it by dividing it by $9$.
$\therefore C = 40^\circ {\text{C}}$
Therefore, the body temperature on the Celsius scale is $40^\circ {\text{C}}$.
Note: As we know that the boiling point temperature of the water in Fahrenheit scale is $212^\circ {\text{F}}$ and in the Celsius scale it is $100^\circ {\text{C}}$ while the freezing point temperature of the water in Fahrenheit scale is $32^\circ {\text{F}}$ and in the Celsius scale it is $0^\circ {\text{C}}$.
Complete step by step solution:
In this question, the body temperature of a patient is \[104^\circ {\text{F}}\]. We need to calculate the body temperature in Celsius scale.
As we know normally available thermometers in the market are calibrated in either Fahrenheit $\left( {^\circ {\text{F}}} \right)$ or degree Celsius $\left( {^\circ {\text{C}}} \right)$ depending on the country, where it is being sold. The United States follows Fahrenheit measurement whereas most other countries follow degree Celsius.
We now know that an increase in temperature corresponds to an increase in the average kinetic energy of atoms and molecules. A result of that increased motion is that the average distance between atoms and molecules increases as the temperature increases. This phenomenon, known as thermal expansion, is the basis for temperature measurement by a liquid thermometer.
Let us assume, the temperature of the scale Fahrenheit $\left( {^\circ {\text{F}}} \right)$ or degree Celsius $\left( {^\circ {\text{C}}} \right)$.
We know that
$9C = 5\left( {F - 32} \right)......\left( 1 \right)$
Where, $C$ is temperature in Celsius and $F$ is in Fahrenheit.
As we know that the temperature that temperature is Fahrenheit is\[104^\circ {\text{F}}\]
We are putting the value of Fahrenheit that is \[104^\circ {\text{F}}\] in equation (1).
$ \Rightarrow 9C = 5\left( {104 - 32} \right)$
Now we subtract the value of Fahrenheit $104$ from $32$ then multiply it by $5$ and simplify it by dividing it by $9$.
$\therefore C = 40^\circ {\text{C}}$
Therefore, the body temperature on the Celsius scale is $40^\circ {\text{C}}$.
Note: As we know that the boiling point temperature of the water in Fahrenheit scale is $212^\circ {\text{F}}$ and in the Celsius scale it is $100^\circ {\text{C}}$ while the freezing point temperature of the water in Fahrenheit scale is $32^\circ {\text{F}}$ and in the Celsius scale it is $0^\circ {\text{C}}$.
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