📘 Chapter 7: Thermal Properties of Matter – Numerical Problems
Prepared by Muhammad Tayyab, Subject Specialist Physics, Govt Christian High School Daska. Based on PECTAA 2026 syllabus (National Curriculum 2023).
📖 What's Inside: This section covers numerical problems from Chapter 7 Thermal Properties of Matter including Temperature Conversion (Celsius, Fahrenheit, Kelvin), Thermometer Scale Reading, and Temperature Equivalence. Each problem is presented with Given Data, To Find, and step-by-step Solution as per the official PECTAA 2026 Physics curriculum. Perfect for Punjab Boards (Lahore, Gujranwala, Multan, etc.) and all BISE boards across Pakistan.
📚 Related Resources – Chapter 7: Thermal Properties of Matter
📋 Exercise Short Qs
📝 Short Qs
📖 Comprehensive Qs
🔢 Numerical Problems
✅ MCQs
✏️ Constructed Response
Thermal Properties of Matter covers temperature, heat, thermometers, temperature scales, and kinetic molecular theory.
📐 Important Formulas – Temperature Conversions
Celsius to Fahrenheit: \(T_F = \frac{9}{5}T_C + 32 = 1.8T_C + 32\)
Fahrenheit to Celsius: \(T_C = \frac{5}{9}(T_F - 32)\)
Celsius to Kelvin: \(T_K = T_C + 273\)
Kelvin to Celsius: \(T_C = T_K - 273\)
Thermometer Scale Reading: \(T_C = \frac{l}{l_{100}} \times 100\)
📐 Numerical Problems & Solutions (PECTAA 2026)
7.1
Normal human body temperature is \(98.6^\circ F\). Convert it into Celsius scale and Kelvin scale.
Given Data
Temperature in Fahrenheit\(T_F = 98.6^\circ F\)
To Find
Temperature in Celsius \(T_C = ?\)
Temperature in Kelvin \(T_K = ?\)
Solution
By using Fahrenheit to Celsius formula:
\[\begin{aligned}
T_C &= \frac{5}{9}(T_F - 32) \\
T_C &= \frac{5}{9}(98.6 - 32) \\
T_C &= \frac{5}{9}(66.6) \\
T_C &= 37^\circ C
\end{aligned}\]
Now by using Celsius to Kelvin formula:
\[\begin{aligned}
T_K &= T_C + 273 \\
T_K &= 37 + 273 \\
T_K &= 310\,K
\end{aligned}\]
✅ \(T_C = 37^\circ C\), \(T_K = 310\,K\)
7.2
At what temperature Celsius and Fahrenheit thermometer reading would be the same?
Given Data
Let temperature in Celsius\(T_C = T\)
And temperature in Fahrenheit\(T_F = T\)
To Find
Temperature at which \(T = T_C = T_F\)
Solution
By using Fahrenheit scale formula:
\[\begin{aligned}
T_F &= \frac{9}{5}T_C + 32 \\
T &= \frac{9}{5}T + 32 \quad (\because T = T_C = T_F) \\
T &= 1.8T + 32 \\
T - 1.8T &= 32 \\
-0.8T &= 32 \\
T &= \frac{32}{-0.8} \\
T &= -40
\end{aligned}\]
✅ \(T = -40^\circ\) (Celsius and Fahrenheit readings are the same at -40°)
7.3
Convert \(5^\circ F\) to Celsius and Kelvin scale.
Given Data
Temperature in Fahrenheit\(T_F = 5^\circ F\)
To Find
Temperature in Celsius \(T_C = ?\)
Temperature in Kelvin \(T_K = ?\)
Solution
By using Fahrenheit scale formula:
\[\begin{aligned}
T_F &= \frac{9}{5}T_C + 32 \\
5 &= 1.8T_C + 32 \\
5 - 32 &= 1.8T_C \\
-27 &= 1.8T_C \\
T_C &= \frac{-27}{1.8} \\
T_C &= -15^\circ C
\end{aligned}\]
Now by using Kelvin scale formula:
\[\begin{aligned}
T_K &= T_C + 273 \\
T_K &= -15 + 273 \\
T_K &= 258\,K
\end{aligned}\]
✅ \(T_C = -15^\circ C\), \(T_K = 258\,K\)
7.4
What is equivalent temperature of \(25^\circ C\) on Fahrenheit and Kelvin scales?
Given Data
Temperature in Celsius\(T_C = 25^\circ C\)
To Find
Temperature in Fahrenheit \(T_F = ?\)
Temperature in Kelvin \(T_K = ?\)
Solution
By using formula of Fahrenheit scale:
\[\begin{aligned}
T_F &= \frac{9}{5}T_C + 32 \\
T_F &= 1.8T_C + 32 \\
T_F &= 1.8(25) + 32 \\
T_F &= 45 + 32 \\
T_F &= 77^\circ F
\end{aligned}\]
Now by using Kelvin scale formula:
\[\begin{aligned}
T_K &= T_C + 273 \\
T_K &= 25 + 273 \\
T_K &= 298\,K
\end{aligned}\]
✅ \(T_F = 77^\circ F\), \(T_K = 298\,K\)
7.5
The ice and steam points on an ungraduated thermometer are found to be \(192\,\text{mm}\) apart. What temperature will be on Celsius scale if the length of mercury thread is at \(67.2\,\text{mm}\) above the ice point mark?
Given Data
Total length between fixed points\(l = 192\,\text{mm}\)
Length above ice point\(l_0 = 67.2\,\text{mm}\)
To Find
Temperature on Celsius scale \(T_C = ?\)
Solution
We use the formula for temperature on a linear scale:
\[\begin{aligned}
T_C &= \frac{l_0}{l} \times 100 \\
T_C &= \frac{67.2}{192} \times 100 \\
T_C &= 35^\circ C
\end{aligned}\]
✅ \(T_C = 35^\circ C\)
7.6
The length between the fixed point of liquid-in-glass thermometer is \(20\,\text{cm}\). If the mercury level is \(4.5\,\text{cm}\) above the lower mark, what is the temperature on the Fahrenheit scale?
Given Data
Total length between fixed points\(l = 20\,\text{cm}\)
Length above lower mark\(l_0 = 4.5\,\text{cm}\)
To Find
Temperature on Fahrenheit scale \(T_F = ?\)
Solution
We use the formula for temperature on a linear scale:
\[\begin{aligned}
T_C &= \frac{l_0}{l} \times 100 \\
T_C &= \frac{4.5}{20} \times 100 \\
T_C &= 22.5^\circ C
\end{aligned}\]
By using Fahrenheit scale formula:
\[\begin{aligned}
T_F &= 1.8T_C + 32 \\
T_F &= (1.8)(22.5) + 32 \\
T_F &= 40.5 + 32 \\
T_F &= 72.5^\circ F
\end{aligned}\]
✅ \(T_F = 72.5^\circ F\)
📘 Solved Examples (PECTAA 2026)
Example 7.1
How much \(30^\circ C\) temperature would be on Fahrenheit and Kelvin scales?
Given Data
Temperature in Celsius\(T_C = 30^\circ C\)
To Find
Temperature in Fahrenheit \(T_F = ?\)
Temperature in Kelvin \(T_K = ?\)
Solution
By using formula of Fahrenheit scale:
\[\begin{aligned}
T_F &= \frac{9}{5}T_C + 32 \\
T_F &= \frac{9}{5}(30) + 32 \\
T_F &= 54 + 32 \\
T_F &= 86^\circ F
\end{aligned}\]
Now by using Kelvin scale formula:
\[\begin{aligned}
T_K &= T_C + 273 \\
T_K &= 30 + 273 \\
T_K &= 303\,K
\end{aligned}\]
✅ \(T_F = 86^\circ F\), \(T_K = 303\,K\)
📐 Key Concepts – Temperature Conversions (Numerical Problems)
Celsius to Fahrenheit: \(T_F = \frac{9}{5}T_C + 32\)
Fahrenheit to Celsius: \(T_C = \frac{5}{9}(T_F - 32)\)
Celsius to Kelvin: \(T_K = T_C + 273\)
Kelvin to Celsius: \(T_C = T_K - 273\)
Thermometer Scale Reading: \(T_C = \frac{l}{l_{100}} \times 100\)
Equivalent Temperature: -40°C = -40°F
💡 Exam Tip:
For numerical problems, always write the given data, the formula being used, and show the steps of your solution clearly. Pay attention to unit conversions and sign conventions (especially for negative temperatures). Practice temperature conversions between Celsius, Fahrenheit, and Kelvin scales thoroughly as they are the foundation for most numericals in this chapter. These problems follow the PECTAA 2026 pattern and are prepared by Subject Specialist Muhammad Tayyab.
Chapter 1Physical Quantities and Measurements
Chapter 2Kinematics
Chapter 3Dynamics
Chapter 4Turning Effect of Forces
Chapter 5Work and Energy
Chapter 6Mechanical Properties of Matter
Chapter 7✨ Thermal Properties of Matter
Chapter 8Magnetism
Chapter 9Nature of Science
📖 Complete syllabus coverage for Class 9 Physics (PECTAA 2026) – Chapters 1 to 9
Created by Hira Science Academy | Aligned with PECTAA 2026 Syllabus