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Chapter 9

Heat and Temperature

NEB Class 11 past exam questions

Past questions and answers

6 questions set from this chapter. Most repeated first.

  • 2082 · HISSAN Central Examination 1011 'A' · 1+1 marks

What is thermal equilibrium? How is it applied in working of mercury thermometer? Explain.

Answer

Thermal equilibrium: two bodies in thermal contact are in thermal equilibrium when there is no net flow of heat between them, which happens when they are at the same temperature.

Application in a mercury thermometer: When the bulb is placed in contact with a body (for example, a patient's mouth), heat flows between the body and the mercury until both reach the same temperature. At that point the mercury and the body are in thermal equilibrium, and the mercury has expanded to a definite length in the capillary. By the zeroth law, the scale reading at this length gives the temperature of the body. The thermometer must therefore be left in contact long enough to reach equilibrium, and its small heat capacity ensures it hardly alters the temperature of the body.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Practice · 1 mark

The coefficient of linear expansion of a solid depends on:
(a) the nature of the material only (b) the original length of the solid only (c) the rise in temperature only (d) the area of cross-section of the solid

Answer

(a) the nature of the material only

The coefficient of linear expansion (α\alpha) is a characteristic property of the material — copper expands more than steel for the same temperature rise. It does not depend on the original length, the temperature rise or the cross-sectional area, since these appear separately in L=L0(1+α Δθ)L = L_0(1 + \alpha\,\Delta\theta).

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Practice · 5 marks

Define heat and temperature. Distinguish between them with two points of difference. Also, explain why temperature remains constant during the change of state of a substance even when heat is supplied.

Answer

Heat is the form of energy that flows from a body at higher temperature to a body at lower temperature because of the temperature difference. It is the total thermal energy of a body (the energy transferred between bodies). SI unit: joule (J).

Temperature is the degree of hotness or coldness of a body. It is a measure of the average kinetic energy of the molecules of the body and decides the direction of heat flow. SI unit: kelvin (K).

Differences

HeatTemperature
A form of energy; SI unit joule (J).A measure of hotness or coldness; SI unit kelvin (K).
Depends on the mass, nature and temperature of the body.Does not depend on the mass of the body.

Why temperature stays constant during a change of state

During melting or boiling, the heat supplied is used to break or loosen the bonds between molecules (increasing their potential energy) and to do work against the external pressure as the volume changes. It does not increase the average kinetic energy of the molecules. Since temperature measures the average kinetic energy, it stays constant until the change of state is complete. This hidden heat is the latent heat.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Practice · 5 marks

Define the coefficient of linear expansion and the coefficient of superficial expansion of a solid. Establish the relation between them: (\beta = 2\alpha), where (\alpha) is the coefficient of linear expansion and (\beta) is the coefficient of superficial expansion.

Answer

Coefficient of linear expansion (α\alpha)

The coefficient of linear expansion of a solid is defined as the fractional increase in length per unit rise in temperature.

If a rod of length ll at temperature θ\theta expands to length l′l' at temperature θ+Δθ\theta + \Delta\theta, then

α=l′−ll Δθ=Δll Δθ\alpha = \frac{l' - l}{l\,\Delta\theta} = \frac{\Delta l}{l\,\Delta\theta}

Coefficient of superficial expansion (β\beta)

The coefficient of superficial expansion of a solid is defined as the fractional increase in surface area per unit rise in temperature.

If a solid sheet of surface area AA at temperature θ\theta expands to area A′A' at temperature θ+Δθ\theta + \Delta\theta, then

β=A′−AA Δθ=ΔAA Δθ\beta = \frac{A' - A}{A\,\Delta\theta} = \frac{\Delta A}{A\,\Delta\theta}

Relation between α\alpha and β\beta

Consider a rectangular sheet of a solid whose length is ll and breadth is bb at temperature θ\theta. Its surface area at this temperature is

A=l bA = l\,b

Let the temperature be raised by Δθ\Delta\theta. If the coefficient of linear expansion of the solid is α\alpha, the new length and breadth are

l′=l(1+α Δθ)l' = l(1 + \alpha\,\Delta\theta) b′=b(1+α Δθ)b' = b(1 + \alpha\,\Delta\theta)

The new surface area is therefore

A′=l′ b′=l(1+α Δθ)  b(1+α Δθ)A' = l'\,b' = l(1 + \alpha\,\Delta\theta)\; b(1 + \alpha\,\Delta\theta) A′=lb (1+α Δθ)2=A (1+α Δθ)2A' = lb\,(1 + \alpha\,\Delta\theta)^2 = A\,(1 + \alpha\,\Delta\theta)^2

Expanding the bracket,

A′=A (1+2α Δθ+α2Δθ2)A' = A\,(1 + 2\alpha\,\Delta\theta + \alpha^2\Delta\theta^2)

Since α\alpha is very small (of the order of 10−510^{-5} per °C), the term α2Δθ2\alpha^2\Delta\theta^2 is negligibly small and can be dropped. Hence

A′=A (1+2α Δθ)A' = A\,(1 + 2\alpha\,\Delta\theta) A′−A=2α A ΔθA' - A = 2\alpha\,A\,\Delta\theta A′−AA Δθ=2α\frac{A' - A}{A\,\Delta\theta} = 2\alpha

But by definition A′−AA Δθ=β\dfrac{A' - A}{A\,\Delta\theta} = \beta. Hence

β=2α\boxed{\beta = 2\alpha}

Thus the coefficient of superficial expansion of a solid is twice its coefficient of linear expansion.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Practice · 8 marks

State the principle of calorimetry. A calorimeter of mass (50,\text{g}) and specific heat capacity (0.4,\text{J g}^{-1},^{\circ}\text{C}^{-1}) contains (100,\text{g}) of water at (20^{\circ}\text{C}). Steam at (100^{\circ}\text{C}) is passed into the calorimeter until the final temperature of the mixture becomes (60^{\circ}\text{C}). Calculate the mass of steam condensed. Given: specific heat capacity of water (= 4.2,\text{J g}^{-1},^{\circ}\text{C}^{-1}) and latent heat of vaporization of steam (= 2260,\text{J g}^{-1}).

Answer

Principle of calorimetry:
When two or more bodies at different temperatures are mixed in an isolated system, the total heat lost by the hotter bodies is equal to the total heat gained by the colder bodies, provided no heat is lost to the surroundings.

[ \text{Heat lost} = \text{Heat gained} ]


Given:

  • Mass of calorimeter, (m_c = 50,\text{g})
  • Specific heat capacity of calorimeter, (c_c = 0.4,\text{J g}^{-1},^{\circ}\text{C}^{-1})
  • Mass of water, (m_w = 100,\text{g})
  • Initial temperature of water and calorimeter, (\theta_1 = 20^{\circ}\text{C})
  • Temperature of steam, (\theta_2 = 100^{\circ}\text{C})
  • Final temperature of mixture, (\theta = 60^{\circ}\text{C})
  • Specific heat capacity of water, (c_w = 4.2,\text{J g}^{-1},^{\circ}\text{C}^{-1})
  • Latent heat of vaporization of steam, (L = 2260,\text{J g}^{-1})

Let the mass of steam condensed be (m_s).


Heat gained by calorimeter and water:

[ Q_{\text{gained}} = (m_c c_c + m_w c_w)(\theta - \theta_1) ]

[ Q_{\text{gained}} = (50 \times 0.4 + 100 \times 4.2)(60 - 20) ]

[ Q_{\text{gained}} = (20 + 420)(40) ]

[ Q_{\text{gained}} = 440 \times 40 = 17600,\text{J} ]


Heat lost by steam:

Steam first condenses at (100^{\circ}\text{C}), then the condensed water cools from (100^{\circ}\text{C}) to (60^{\circ}\text{C}).

[ Q_{\text{lost}} = m_s L + m_s c_w (100 - 60) ]

[ Q_{\text{lost}} = m_s(2260 + 4.2 \times 40) ]

[ Q_{\text{lost}} = m_s(2260 + 168) ]

[ Q_{\text{lost}} = 2428,m_s ]


By the principle of calorimetry:

[ Q_{\text{lost}} = Q_{\text{gained}} ]

[ 2428,m_s = 17600 ]

[ m_s = \frac{17600}{2428} ]

[ m_s = 7.25,\text{g} ]


Answer:
The mass of steam condensed is approximately

[ \boxed{7.25,\text{g}} ]

Answer checked by Claude Sonnet · Oct 21, 2026

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  • NEB Grade XI Physics Sub.Code 110 · 2 marks

The triple point of water is chosen as a standard fixed point in modern thermometry. Why?

Answer

The triple point of water is the unique condition (T=273.16 KT=273.16\ \text{K}, P=611.7 PaP=611.7\ \text{Pa}) at which ice, liquid water and water vapour coexist in equilibrium. It occurs at exactly one temperature and one pressure, so it can be reproduced precisely in any laboratory with a sealed triple-point cell. The ice and steam points, by contrast, depend on the pressure and on impurities, so they are less sharp and less reproducible. This makes the triple point a reliable standard fixed point.

Answer checked by Claude Sonnet · Oct 21, 2026

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Questions from CDC Class 11 Physics Model Question (Specification Grid 2077) (CDC model paper, Grade 11, 2077), NEB Class 11 Physics Model Question 1011 (2077/78) (NEB/CDC model paper, Grade 11, 2077/78), Practice questions written for this site (Written from the official syllabus and specification grid where no past papers cover a chapter), NEB Grade XI 2074 (2017) Physics Sub.Code 110 'A' (NEB board exam, 2074 (2017), old course. Only questions that fit a chapter of the current outline were kept.), NEB Grade XI 2076 (2019) Physics Sub.Code 110 'B' (NEB board exam, 2076 (2019), old course (last regular old-course Class 11 paper). Only questions that fit a chapter of the current outline were kept.), NEB Grade XI Physics Sub.Code 110 (old course, past annual paper, year not stated) (NEB's own sample of a past annual paper from the old course (folder 'Class 11 Model Question, old course'). Only questions that fit a chapter of the current outline were kept.), HISSAN Central Examination 2082 (2025) Class XI Physics 1011 'A' (Joint school-association exam (HISSAN), 2082 (2025), not an NEB paper. Questions read from the paper PDF; answers not used.), Class 11 Physics model question Set 1 (new syllabus, 2022) (Practice/model set 1 posted on gaurab123.com.np in 2022 (written in CDC model-paper style, not a board paper). Only the questions were taken; obvious typos fixed.), Class 11 Physics model question Set 2 (new syllabus, 2022) (Practice/model set 2 posted on gaurab123.com.np in 2022. Only the questions were taken; the page's printed answers were not used.) and Class 11 Physics model question Set 3 (new syllabus, 2022) (Practice/model set 3 posted on gaurab123.com.np in 2022. Only the questions were taken; the page's printed answers were not used.). Answers are written for this site; check them against your class notes.

Chapter titles and hours from the official syllabus ↗