
The body temperature of a patient is \[104^\circ {\text{F}}\]. Calculate his body temperature in Celsius scale $?$
Answer
228.3k+ views
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 $0^\circ {\text{C}}$.
Recently Updated Pages
States of Matter Chapter For JEE Main Chemistry

Circuit Switching vs Packet Switching: Key Differences Explained

Mass vs Weight: Key Differences Explained for Students

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Isoelectronic Definition in Chemistry: Meaning, Examples & Trends

Ionisation Energy and Ionisation Potential Explained

Trending doubts
JEE Main 2026: Admit Card Out, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Laws of Motion Class 11 Physics Chapter 4 CBSE Notes - 2025-26

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

Waves Class 11 Physics Chapter 14 CBSE Notes - 2025-26

Thermodynamics Class 11 Physics Chapter 11 CBSE Notes - 2025-26

