🔥 Chapter 10: Thermal Physics – 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 10 Thermal Physics including volume expansion, linear expansion, specific heat capacity, latent heat of fusion, and latent heat of vaporization. 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 10: Thermal Physics
Thermal Physics covers temperature, heat, thermal expansion, specific heat capacity, latent heat, and thermal equilibrium. Includes solved examples and numerical problems.
📐 Numerical Problems & Solutions (PECTAA 2026)
10.1 Coefficient of Linear Expansion
A metal rod of length 1 m expands by 0.02 m when heated from 20°C to 120°C. Calculate its coefficient of linear expansion.
Given Data
\(Original\ length\)\(= L_o = 1 \, \text{m}\)
\(Change\ in\ length\)\(= \Delta L = 0.02 \, \text{m}\)
\(Initial\ temperature\)\(= T_o = 20^\circ \text{C} = 293 \, \text{K}\)
\(Final\ temperature\)\(= T = 120^\circ \text{C} = 393 \, \text{K}\)
\(Change\ in\ temperature\)\(= \Delta T = 393 - 293 = 100 \, \text{K}\)
To Find
\(Coefficient\ of\ linear\ expansion = \alpha = ?\)
Solution
By using formula of coefficient of linear expansion:
\[\begin{aligned}
\alpha &= \frac{\Delta L}{L_o \Delta T} \\
\alpha &= \frac{0.02}{(1)(100)} \\
\boldsymbol{\alpha} &= \boldsymbol{2 \times 10^{-4} \, \text{K}^{-1}}
\end{aligned}\]
✅ \(\alpha = 2.0 \times 10^{-4} \, \text{K}^{-1}\)
10.2 Volume Expansion of Water
A container holds 1 litre of water at 20°C. What will be its volume at 80°C assuming water's coefficient of volume expansion is \(2.1 \times 10^{-4}\) per °C?
Given Data
\(Original\ volume\)\(= V_o = 1 \, \text{litre} = 1000 \, \text{cm}^3\)
\(Initial\ temperature\)\(= T_o = 20^\circ \text{C}\)
\(Final\ temperature\)\(= T = 80^\circ \text{C}\)
\(Change\ in\ temperature\)\(= \Delta T = 80 - 20 = 60^\circ \text{C}\)
\(Coefficient\ of\ volume\ expansion\)\(= \beta = 2.1 \times 10^{-4} \, \text{per °C}\)
To Find
\(Final\ volume = V = ?\)
Solution
By using formula of volume thermal expansion:
\[\begin{aligned}
V &= V_o (1 + \beta \Delta T) \\
V &= 1000 [1 + (2.1 \times 10^{-4})(60)] \\
V &= 1000 [1 + 0.0126] \\
V &= 1000 [1.0126] \\
\boldsymbol{V} &= \boldsymbol{1012.6 \, \text{cm}^3}
\end{aligned}\]
✅ \(V = 1012.6 \, \text{cm}^3\)
10.3 Length of Steel Rod at Higher Temperature
A steel rod initially measures 2 m at 20°C. If its coefficient of linear expansion is \(1.2 \times 10^{-5} \, \text{°C}^{-1}\), what will be its length at 100°C?
Given Data
\(Original\ length\)\(= L_o = 2 \, \text{m}\)
\(Initial\ temperature\)\(= T_o = 20^\circ \text{C}\)
\(Final\ temperature\)\(= T = 100^\circ \text{C}\)
\(Change\ in\ temperature\)\(= \Delta T = 100 - 20 = 80^\circ \text{C}\)
\(Coefficient\ of\ linear\ expansion\)\(= \alpha = 1.2 \times 10^{-5} \, \text{°C}^{-1}\)
To Find
\(Final\ length = L = ?\)
Solution
By using formula of linear thermal expansion:
\[\begin{aligned}
L &= L_o (1 + \alpha \Delta T) \\
L &= 2 [1 + (1.2 \times 10^{-5})(80)] \\
L &= 2 [1 + 9.6 \times 10^{-4}] \\
L &= 2 [1.00096] \\
\boldsymbol{L} &= \boldsymbol{2.00192 \, \text{m}}
\end{aligned}\]
✅ \(L = 2.00192 \, \text{m}\)
10.4 Temperature Change from Bridge Expansion
A steel bridge expands by 5 cm on a hot summer day. If the bridge originally spanned 100 m, what is the temperature change? (\(\alpha = 1.2 \times 10^{-5} \, \text{°C}^{-1}\))
Given Data
\(Original\ span\)\(= L_o = 100 \, \text{m}\)
\(Change\ in\ length\)\(= \Delta L = 5 \, \text{cm} = 0.05 \, \text{m}\)
\(Coefficient\ of\ linear\ expansion\)\(= \alpha = 1.2 \times 10^{-5} \, \text{°C}^{-1}\)
To Find
\(Temperature\ change = \Delta T = ?\)
Solution
By using formula of linear expansion:
\[\begin{aligned}
\Delta L &= \alpha L_o \Delta T \\
\Delta T &= \frac{\Delta L}{\alpha L_o} \\
\Delta T &= \frac{0.05}{(1.2 \times 10^{-5})(100)} \\
\Delta T &= \frac{0.05}{1.2 \times 10^{-3}} \\
\boldsymbol{\Delta T} &= \boldsymbol{41.67^\circ \text{C}}
\end{aligned}\]
✅ \(\Delta T = 41.67^\circ \text{C}\)
10.5 Heat Required to Raise Temperature of Iron Bar
How much heat is required to raise the temperature of a 2 kg iron bar from 20°C to 100°C given that the specific heat capacity of iron is \(450 \, \text{J kg}^{-1} \text{K}^{-1}\)?
Given Data
\(Mass\ of\ iron\ bar\)\(= m = 2 \, \text{kg}\)
\(Initial\ temperature\)\(= T_o = 20^\circ \text{C} = 293 \, \text{K}\)
\(Final\ temperature\)\(= T = 100^\circ \text{C} = 373 \, \text{K}\)
\(Change\ in\ temperature\)\(= \Delta T = 373 - 293 = 80 \, \text{K}\)
\(Specific\ heat\ of\ iron\)\(= c = 450 \, \text{J kg}^{-1} \text{K}^{-1}\)
To Find
\(Amount\ of\ heat = Q = ?\)
Solution
By using formula of heat transfer:
\[\begin{aligned}
Q &= c m \Delta T \\
Q &= (450)(2)(80) \\
Q &= 72000 \, \text{J} \\
\boldsymbol{Q} &= \boldsymbol{72 \, \text{kJ}}
\end{aligned}\]
✅ \(Q = 72 \, \text{kJ}\)
10.6 Latent Heat of Fusion (Ice Melting)
How much heat is required to melt 500 g of ice at 0°C into water at 0°C? (Latent heat of fusion of ice \(= 3.36 \times 10^5 \, \text{J kg}^{-1}\))
Given Data
\(Mass\ of\ ice\)\(= m = 500 \, \text{g} = 0.5 \, \text{kg}\)
\(Latent\ heat\ of\ fusion\)\(= L_f = 3.36 \times 10^5 \, \text{J kg}^{-1}\)
To Find
\(Heat\ required\ for\ melting = Q = ?\)
Solution
By using formula of latent heat of fusion:
\[\begin{aligned}
Q &= m L_f \\
Q &= (0.5)(3.36 \times 10^5) \\
Q &= 168000 \, \text{J} \\
\boldsymbol{Q} &= \boldsymbol{168 \, \text{kJ}}
\end{aligned}\]
✅ \(Q = 168 \, \text{kJ}\)
10.7 Latent Heat of Vaporization
Calculate the heat required to completely vaporize 1 kg of water at 100°C. (Latent heat of vaporization of water \(= 2.26 \times 10^6 \, \text{J kg}^{-1}\))
Given Data
\(Mass\ of\ water\)\(= m = 1 \, \text{kg}\)
\(Latent\ heat\ of\ vaporization\)\(= L_v = 2.26 \times 10^6 \, \text{J kg}^{-1}\)
To Find
\(Heat\ required\ for\ vaporization = Q = ?\)
Solution
By using formula of latent heat of vaporization:
\[\begin{aligned}
Q &= m L_v \\
Q &= (1)(2.26 \times 10^6) \\
Q &= 2.26 \times 10^6 \, \text{J} \\
\boldsymbol{Q} &= \boldsymbol{2.26 \, \text{MJ}}
\end{aligned}\]
✅ \(Q = 2.26 \, \text{MJ}\)
📘 Examples with Solutions
Ex 10.1 Coefficient of Linear Expansion
A metal rod of length 1.5 m expands by 0.025 m when heated from 30°C to 180°C. Calculate its coefficient of linear expansion.
Given Data
\(Original\ length\)\(= L_o = 1.5 \, \text{m}\)
\(Change\ in\ length\)\(= \Delta L = 0.025 \, \text{m}\)
\(Initial\ temperature\)\(= T_o = 30^\circ \text{C} = 303 \, \text{K}\)
\(Final\ temperature\)\(= T = 180^\circ \text{C} = 453 \, \text{K}\)
\(Change\ in\ temperature\)\(= \Delta T = 453 - 303 = 150 \, \text{K}\)
To Find
\(Coefficient\ of\ linear\ expansion = \alpha = ?\)
Solution
By using formula of coefficient of linear expansion:
\[\begin{aligned}
\alpha &= \frac{\Delta L}{L_o \Delta T} \\
\alpha &= \frac{0.025}{(1.5)(150)} \\
\alpha &= \frac{0.025}{225} \\
\boldsymbol{\alpha} &= \boldsymbol{1.11 \times 10^{-4} \, \text{K}^{-1}}
\end{aligned}\]
✅ \(\alpha = 1.11 \times 10^{-4} \, \text{K}^{-1}\)
Ex 10.2 Heat Required to Melt Ice
How much heat is required to melt 600 g of ice at 0°C into water at 0°C? Latent heat of fusion of ice is \(3.36 \times 10^5 \, \text{J kg}^{-1}\).
Given Data
\(Mass\ of\ ice\)\(= m = 600 \, \text{g} = 0.6 \, \text{kg}\)
\(Latent\ heat\ of\ fusion\)\(= L_f = 3.36 \times 10^5 \, \text{J kg}^{-1}\)
To Find
\(Heat\ required\ for\ melting = Q = ?\)
Solution
By using formula of latent heat of fusion:
\[\begin{aligned}
Q &= m L_f \\
Q &= (0.6)(3.36 \times 10^5) \\
Q &= 201600 \, \text{J} \\
\boldsymbol{Q} &= \boldsymbol{201.6 \, \text{kJ}}
\end{aligned}\]
✅ \(Q = 201.6 \, \text{kJ}\)
📐 Key Formulas – Thermal Physics
Linear Expansion: \( \Delta L = \alpha L_o \Delta T \) or \( L = L_o (1 + \alpha \Delta T) \)
Volume Expansion: \( \Delta V = \beta V_o \Delta T \) or \( V = V_o (1 + \beta \Delta T) \)
Heat Transfer: \( Q = c m \Delta T \)
Latent Heat of Fusion: \( Q = m L_f \)
Latent Heat of Vaporization: \( Q = m L_v \)
Coefficient of Linear Expansion: \( \alpha = \frac{\Delta L}{L_o \Delta T} \)
Coefficient of Volume Expansion: \( \beta = \frac{\Delta V}{V_o \Delta T} \)
Unit 10✨ Thermal Physics
Unit 11Transfer of Thermal Energy
Unit 12Waves
Unit 13Sound
Unit 14Light
Unit 15Electrostatics
Unit 16Electricity
Unit 17Electromagnetism
Unit 18Electromagnetic Induction & EM Waves
Unit 19Electronics
Unit 20Atomic and Nuclear Physics
Unit 21Space and Environment
📖 Complete syllabus coverage for Class 10 Physics (PECTAA 2026) – Units 10 to 21
💡 Exam Tip:
For numerical problems, always write Given Data, To Find, and step-by-step Solution with formulas. Pay attention to units and temperature conversions (K vs °C). These problems follow the PECTAA 2026 pattern and are prepared by Subject Specialist Muhammad Tayyab.
Created by Hira Science Academy | Aligned with PECTAA 2026 Syllabus