PiTheory

PiTheory JEE Physics Practice & Revision Pack · Class 11 · Chapter 10 · JEE Main

Thermal Properties of Matter

Chapter-wise practice and revision for JEE Main: concise notes on temperature scales, thermal expansion, calorimetry, conduction, radiation and Newton's law of cooling, concept figures, and 40 selected previous-year JEE Main questions covering the major concepts and question patterns, each with a step-by-step solution.

40 selected questions4 topics4 concept figuresPreview · first 3 of 40 questions

How to study this chapter

  1. Look at the topic's figures, then read its notes (5 to 10 minutes). They give the major equations and methods the questions use.
  2. Solve the topic's questions with a timer: 2 minutes for Easy, 3 to 4 for Medium and Hard.
  3. Check the solution. Read the Point to note line even if you were right: it is the idea you will reuse.
  4. Mark each question Got it or Revise. Come back to the Revise list after two days.

Where the marks come from

Main idea of each of the 89 previous-year JEE questions in the PiTheory bank

Calorimetry34
Scales, expansion24
Conduction16
Radiation, cooling13

Contents

  1. 1Temperature scales and thermal expansionQ1–Q11
  2. 2Specific heat, calorimetry and latent heatQ12–Q24
  3. 3Thermal conductionQ25–Q32
  4. 4Radiation and Newton's law of coolingQ33–Q40
  5. ✓Last-day checklist
01

Topic 1 · Q1–Q11

Temperature scales and thermal expansion

24 of the 89 JEE PYQs in the bank test this as their main idea

Look first

brass (larger α)ironcold: straighthot: bends, brass outsideΔL = Lα ΔTΔA = A(2α) ΔTΔV = V(3α) ΔTa hole grows likethe metal around it
Different metals expand by different amounts.

Quick notes

$$\frac{C}{100}=\frac{F-32}{180}=\frac{K-273}{100}=\frac{X-X_{\text{ice}}}{X_{\text{steam}}-X_{\text{ice}}}$$

Every linear scale is the same fraction of the ice-to-steam interval. For temperature differences, only the size of a degree matters: $\Delta F=\tfrac95\Delta C$, $\Delta K=\Delta C$. The Celsius and Fahrenheit scales agree at $-40^\circ$.

$$\Delta L=L\alpha\Delta T,\qquad \Delta A=A\beta\Delta T,\qquad \Delta V=V\gamma\Delta T,\qquad \alpha:\beta:\gamma=1:2:3$$

Holes and cavities expand like the material around them. Density falls: $\dfrac{\Delta\rho}{\rho}=-\gamma\Delta T$. Anisotropic solid: $\gamma=\alpha_x+\alpha_y+\alpha_z$.

SituationResult
Rod held between rigid wallsstress $Y\alpha\Delta T$, force $YA\alpha\Delta T$ (independent of length)
Two rods, difference in length constant$L_1\alpha_1=L_2\alpha_2$
Empty space in a vessel constant$V_{\text{vessel}}\gamma_{\text{vessel}}=V_{\text{liquid}}\gamma_{\text{liquid}}$
Liquid in a vesselapparent expansion $\gamma_{\text{app}}=\gamma_{\text{liquid}}-\gamma_{\text{vessel}}$
Pendulum clock heatedloses time; fractional loss per second $\tfrac12\alpha\Delta T$
Rods joined end to end$\alpha_{\text{eff}}=\dfrac{\alpha_1L_1+\alpha_2L_2}{L_1+L_2}$
Restore a compressed cube by heating$\dfrac PK=3\alpha\Delta T$
Traps. Use $\alpha$, $2\alpha$ or $3\alpha$ to match the dimension asked. A clamped wire that is cooled goes into tension; a clamped rod that is heated goes into compression.

Practice: first 3 of 11 questions

Q1JEE Main 2024 (4 Apr, Shift 1)Easy

On the Celsius scale the temperature of a body increases by $40^\circ\text{C}$. The increase in temperature on the Fahrenheit scale is

  1. A$68^\circ\text{F}$
  2. B$75^\circ\text{F}$
  3. C$72^\circ\text{F}$
  4. D$70^\circ\text{F}$
Answer and solution

Answer: Option C: $72^\circ\text{F}$

Concept. A temperature *change* converts with the scale factor only: $100$ Celsius degrees span the same interval as $180$ Fahrenheit degrees, so $\Delta F=\dfrac95\Delta C$. The offset $32$ cancels in a difference.

Step 1. Convert the change. $\Delta F=\dfrac95\times40=72$.

Answer. $72^\circ$F (Option C).

Point to note. Differences: $\Delta F=\frac95\Delta C$.

Q2JEE Main 2023 (30 Jan, Shift 2)Easy

A faulty thermometer reads $5^\circ\text{C}$ in melting ice and $95^\circ\text{C}$ in steam. The correct temperature on absolute scale will be ______ K when the faulty thermometer reads $41^\circ\text{C}$.

Numerical answer type

Answer and solution

Answer: 313

Concept. Any linear scale is compared through the fraction of the ice-to-steam interval: $\dfrac{\text{reading}-\text{ice point}}{\text{steam point}-\text{ice point}}$ is the same on every scale.

Step 1. Fraction on the faulty scale. $\dfrac{41-5}{95-5}=\dfrac{36}{90}=0.4$.

Step 2. Celsius value. $0.4\times100=40^\circ$C.

Step 3. Kelvin. $40+273=313$ K.

Answer. $313$.

Point to note. Same fraction of the ice-to-steam interval.

Q3JEE Main 2023 (25 Jan, Shift 2)Medium

The graph between two temperature scales $P$ and $Q$ is shown in the figure. Between upper fixed point and lower fixed point there are $150$ equal divisions of scale $P$ and $100$ divisions on scale $Q$. The relationship for conversion between the two scales is given by

Straight-line graph of temperature on scale P against scale Q: from P = 30 at Q = 0 to P = 180 at Q = 100
  1. A$\dfrac{t_Q}{150}=\dfrac{t_P-180}{100}$
  2. B$\dfrac{t_Q}{100}=\dfrac{t_P-30}{150}$
  3. C$\dfrac{t_P}{180}=\dfrac{t_Q-40}{100}$
  4. D$\dfrac{t_Q}{100}=\dfrac{t_P-180}{150}$
Answer and solution

Answer: Option B: $\dfrac{t_Q}{100}=\dfrac{t_P-30}{150}$

Concept. Two linear scales are related by matching the same fraction of the fixed-point interval. From the graph, scale $Q$ runs $0$ to $100$ while scale $P$ runs from its lower fixed point up to $180$.

Step 1. Lower fixed point on $P$. There are $150$ divisions ending at $180$, so it starts at $180-150=30$.

Step 2. Equal fractions. $$\frac{t_Q-0}{100}=\frac{t_P-30}{150}$$

Answer. Option B.

Point to note. Lower fixed point of $P$ is $180-150$.

37 more questions and 3 more topics in the full chapter

  1. 2Specific heat, calorimetry and latent heat13 questions with solutions
  2. 3Thermal conduction8 questions with solutions
  3. 4Radiation and Newton's law of cooling8 questions with solutions
  4. ✓Last-day checklist

The complete Thermal Properties of Matter chapter is in the Class 11 Pack and the Complete Pack, as a PDF on WhatsApp. Want to see a whole chapter first? Laws of Motion and Current Electricity are free in full.

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