IOE entrance Physics · Chapter 2
Heat and Thermodynamics
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51 questions in 5 syllabus topics · 20 are 2-mark questions.
2.1 Temperature and quantity of heat
11 questions
1. Two ice blocks when pressed together join to form one block, because
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Think of regelation.
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Answer: B. melting point falls with pressure
Pressure lowers the melting point of ice, so the ice at the contact melts; when the pressure is released the water refreezes and joins the blocks (regelation).
2. The triple point of water, used as the fixed point of the Kelvin scale in the NEB syllabus, is at a temperature of
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The triple point is not the same as the normal melting point.
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Answer: A. 273.16 K
The triple point of water (where ice, water and vapour coexist) is at 273.16 K (0.01 °C). 273.15 K is the ice point at 1 atm (0 °C), slightly lower. (Since 2019 the kelvin is formally defined through the Boltzmann constant, but the triple point is still 273.16 K to high accuracy.)
3. The zeroth law of thermodynamics provides the basis for
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Think about what bodies in thermal equilibrium have in common.
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Answer: D. the concept of temperature
If A and B are each in thermal equilibrium with C, they are in equilibrium with each other. This shared property is temperature, so the zeroth law underlies thermometry.
4. 10 g of ice at 0 °C is dropped into 50 g of water at 40 °C in an insulated container. Final temperature? (Latent heat of fusion = 336 J/g, specific heat of water = 4.2 J g⁻¹ °C⁻¹)
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First check whether all the ice melts, then share the leftover heat among the total mass.
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Answer: D. 20 °C
Heat to melt ice = 10 × 336 = 3360 J. Heat released by water cooling to 0 °C = 50 × 4.2 × 40 = 8400 J. Remaining 5040 J heats all 60 g: ΔT = 5040/(60 × 4.2) = 20 °C.
5. Newton's law of cooling states that the rate of loss of heat from a body is proportional to
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Newton's law is a linear law.
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Answer: C. the excess of its temperature over the surroundings, for small temperature differences
dT/dt = −k(T − T₀), valid when T − T₀ is small. The T⁴ dependence belongs to Stefan's law.
6. A body cools from 70 °C to 50 °C in 5 minutes in a room at 20 °C. Using Newton's law of cooling (average-temperature form), the time taken to cool from 50 °C to 40 °C is
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Use (T₁ − T₂)/t = k[(T₁ + T₂)/2 − T₀] for both intervals.
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Answer: A. 4 min
First interval: 20/5 = k(60 − 20) gives k = 0.1 per min. Second interval: 10/t = 0.1 × (45 − 20) = 2.5, so t = 4 min.
7. At what temperature do a Celsius thermometer and a Fahrenheit thermometer show the same numerical reading?
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Put F = C in the conversion formula.
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Answer: A. −40°
F = (9/5)C + 32. Setting F = C: C − 9C/5 = 32, so −4C/5 = 32 and C = −40.
8. Heat needed to change 1 g of ice at −10 °C completely into steam at 100 °C is (specific heat of ice 2.1 J g⁻¹ °C⁻¹, of water 4.2 J g⁻¹ °C⁻¹, Lf = 336 J/g, Lv = 2260 J/g)
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Add the heat for each of the four stages.
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Answer: D. 3037 J
Ice −10 → 0 °C: 21 J; melting: 336 J; water 0 → 100 °C: 420 J; vaporising: 2260 J. Total = 21 + 336 + 420 + 2260 = 3037 J.
9. Two solid spheres of the same material have radii r and 2r. The ratio of their heat capacities is
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Heat capacity depends on mass, not just material.
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Answer: C. 1 : 8
Heat capacity = mc, and mass ∝ volume ∝ r³. So the ratio is r³ : (2r)³ = 1 : 8. Specific heat (per kg) would be the same for both.
10. While a solid is melting at its melting point, the temperature stays constant because the heat supplied
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Temperature is linked to kinetic energy, not potential energy.
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Answer: C. increases the potential energy of the molecules by breaking their bonds
Temperature measures average molecular kinetic energy. During melting the latent heat goes into increasing intermolecular potential energy, so kinetic energy and temperature do not change.
11. Water falls from a height of 420 m. If all its potential energy is converted into heat that stays in the water, the rise in temperature is (g = 10 m/s², specific heat of water = 4200 J kg⁻¹ K⁻¹)
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Equate mgh to mcΔT.
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Answer: A. 1 °C
mgh = mcΔT, so ΔT = gh/c = (10 × 420)/4200 = 1 °C. The mass cancels.
2.2 Thermal expansion
10 questions
12. For an isotropic solid with coefficients of linear (α), superficial (β) and cubical (γ) expansion,
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Expand (1 + αΔT)² and (1 + αΔT)³, keeping only first-order terms.
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Answer: B. β = 2α and γ = 3α
Area A(1 + αΔT)² ≈ A(1 + 2αΔT) and volume V(1 + αΔT)³ ≈ V(1 + 3αΔT), so α : β : γ = 1 : 2 : 3.
13. A steel rod 2 m long (α = 1.2×10⁻⁵ °C⁻¹) is heated from 20 °C to 70 °C. Its increase in length is
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Use ΔL = LαΔT.
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Answer: B. 1.2 mm
ΔL = LαΔT = 2 × 1.2×10⁻⁵ × 50 = 1.2×10⁻³ m = 1.2 mm.
14. A steel rod of cross-section 2 cm² is clamped rigidly at both ends at 20 °C and then heated to 70 °C. Force exerted on the clamps? (Y = 2×10¹¹ Pa, α = 1.2×10⁻⁵ °C⁻¹)
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The rod is stopped from expanding, so strain = αΔT. Then use stress = Y × strain.
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Answer: B. 2.4×10⁴ N
Thermal stress = YαΔT = 2×10¹¹ × 1.2×10⁻⁵ × 50 = 1.2×10⁸ Pa. Force = stress × area = 1.2×10⁸ × 2×10⁻⁴ = 2.4×10⁴ N.
15. Water has its maximum density at
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Recall the anomalous expansion of water.
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Answer: C. 4 °C
Because of its anomalous expansion, water contracts when heated from 0 °C to 4 °C and expands above 4 °C. Its density is therefore greatest at 4 °C.
16. A liquid in a glass vessel has an apparent coefficient of cubical expansion of 1.5×10⁻⁴ °C⁻¹. The coefficient of linear expansion of the glass is 1×10⁻⁵ °C⁻¹. The real coefficient of cubical expansion of the liquid is
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The vessel expands too. Use its cubical coefficient, not its linear one.
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Answer: B. 1.8×10⁻⁴ °C⁻¹
γ_real = γ_apparent + γ_vessel, and γ_vessel = 3α = 3×10⁻⁵ °C⁻¹. So γ_real = 1.5×10⁻⁴ + 0.3×10⁻⁴ = 1.8×10⁻⁴ °C⁻¹.
17. A bimetallic strip made of brass and iron is straight at room temperature. When it is heated, it bends so that
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Which arc of a curve is longer, the inner one or the outer one?
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Answer: B. the metal with the larger expansion coefficient is on the convex (outer) side
The metal that expands more becomes longer, so it takes the longer outer arc. This places it on the convex side.
18. A pendulum clock with a brass pendulum (α = 2×10⁻⁵ °C⁻¹) keeps correct time at 20 °C. At 30 °C, the clock will
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The fractional change in period is half the fractional change in length.
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Answer: D. lose about 8.64 s per day
T ∝ √L, so ΔT/T = ½αΔθ = ½ × 2×10⁻⁵ × 10 = 10⁻⁴. The time lost per day = 10⁻⁴ × 86400 = 8.64 s. A longer period means the clock runs slow, so it loses time.
19. When a substance with cubical expansion coefficient γ is heated through a small ΔT, the fractional change in its density is approximately
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Density varies inversely with volume.
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Answer: B. −γΔT
ρ' = ρ/(1 + γΔT) ≈ ρ(1 − γΔT), so Δρ/ρ ≈ −γΔT. The density decreases as the volume increases.
20. A metal plate has a circular hole in it. When the plate is heated uniformly, the diameter of the hole
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Think of thermal expansion as a uniform photographic enlargement.
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Answer: A. increases, as if the hole were made of the same metal
Uniform heating scales every length in the plate by (1 + αΔT), including the hole's diameter. The hole therefore expands like a disc of the same metal.
21. A glass flask of volume 500 cm³ at 0 °C is completely filled with mercury at 0 °C and then heated to 100 °C. How much mercury overflows? (γ of mercury = 1.8×10⁻⁴ °C⁻¹, α of glass = 9×10⁻⁶ °C⁻¹)
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Overflow equals the mercury's expansion minus the flask's expansion.
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Answer: B. 7.65 cm³
γ_glass = 3α = 2.7×10⁻⁵ °C⁻¹. Overflow = V(γ_Hg − γ_glass)ΔT = 500 × (1.8×10⁻⁴ − 0.27×10⁻⁴) × 100 = 500 × 1.53×10⁻² = 7.65 cm³.
2.3 Transfer of heat
10 questions
22. If the absolute temperature of a black body is doubled, the total power it radiates becomes
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P ∝ T⁴.
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Answer: B. 16 times
By the Stefan–Boltzmann law, P = σAT⁴. Doubling T multiplies P by 2⁴ = 16.
23. The spectra of two stars A and B peak at wavelengths of 400 nm and 600 nm. The ratio of their surface temperatures T_A : T_B is
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Peak wavelength is inversely proportional to temperature.
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Answer: A. 3 : 2
Wien's law gives λ_m T = constant, so T_A/T_B = λ_B/λ_A = 600/400 = 3/2.
24. Two slabs of equal thickness and area are placed in contact. Their thermal conductivities are 2k and k. The free face of the 2k slab is at 100 °C and the free face of the k slab is at 0 °C. In steady state, the temperature of the interface is about
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Equate the heat flow rates through the two slabs.
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Answer: B. 66.7 °C
In steady state the heat current is equal in both slabs: 2k(100 − T)/d = k(T − 0)/d. This gives 200 − 2T = T, so T = 200/3 ≈ 66.7 °C. The better conductor has the smaller temperature drop.
25. A perfect black body is one that
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Think of absorptivity, not colour.
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Answer: D. absorbs all radiation falling on it, at every wavelength
A perfect black body has absorptivity 1 at all wavelengths. By Kirchhoff's law it is also the best possible emitter (emissivity 1), so it does emit radiation.
26. Two slabs of equal thickness and equal area, with thermal conductivities 3 W m⁻¹ K⁻¹ and 6 W m⁻¹ K⁻¹, are joined face to face in series. The equivalent thermal conductivity of the combination is
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Add thermal resistances, d/(kA), for slabs in series.
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Answer: C. 4 W m⁻¹ K⁻¹
For series slabs, total thermal resistance = sum of resistances: 2d/(k_eq A) = d/(3A) + d/(6A). So k_eq = 2k₁k₂/(k₁ + k₂) = 2 × 18/9 = 4 W m⁻¹ K⁻¹. The value 4.5 is the parallel (arithmetic mean) result.
27. Natural convection currents cannot be set up
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What force makes hot fluid rise?
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Answer: B. inside a spacecraft orbiting freely in a weightless condition
Natural convection depends on buoyancy: warmer, less dense fluid rises under gravity. With no effective gravity, as in free-fall orbit, buoyant forces disappear and convection currents do not form.
28. Two spheres made of the same material have radii in the ratio 1 : 2 and absolute temperatures in the ratio 2 : 1. The ratio of the powers they radiate is
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Radiated power depends on both surface area and T⁴.
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Answer: A. 4 : 1
P ∝ AT⁴ ∝ r²T⁴. Ratio = (1² × 2⁴) : (2² × 1⁴) = 16 : 4 = 4 : 1.
29. According to Kirchhoff's law of radiation, at a given temperature and wavelength, the ratio of a body's emissive power to its absorptive power
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Good absorbers are good emitters.
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Answer: A. is the same for all bodies and equals the emissive power of a black body
e_λ/a_λ = E_λ, which is the same for every body at a given temperature and wavelength. So good absorbers are also good emitters.
30. The solar spectrum peaks at about 500 nm. Taking Wien's constant b = 2.9×10⁻³ m K, the Sun's surface temperature is about
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Use λ_m T = b.
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Answer: D. 5800 K
T = b/λ_m = 2.9×10⁻³ / 5×10⁻⁷ = 5.8×10³ K.
31. A black body at 400 K in surroundings at 200 K loses net power P by radiation. If the body's temperature is raised to 600 K, with the surroundings unchanged, the net power lost becomes
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Use T⁴ − T₀⁴, not T⁴ alone.
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Answer: C. 16P/3
Net power ∝ T⁴ − T₀⁴. At 400 K: 200⁴(2⁴ − 1) = 15 × 200⁴. At 600 K: 200⁴(3⁴ − 1) = 80 × 200⁴. Ratio = 80/15 = 16/3.
2.4 Kinetic theory and heat capacities
10 questions
32. At the same temperature, the ratio of the rms speed of hydrogen (H₂) molecules to that of oxygen (O₂) molecules is
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At a fixed T, v_rms ∝ 1/√M.
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Answer: A. 4 : 1
v_rms = √(3RT/M) ∝ 1/√M. Ratio = √(32/2) = √16 = 4.
33. The rms speed of the molecules of a gas at 27 °C is v. At what temperature will the rms speed be 2v?
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Convert to kelvin first. T ∝ v².
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Answer: C. 927 °C
v_rms ∝ √T, so doubling the speed needs 4 times the absolute temperature: 4 × 300 K = 1200 K = 927 °C.
34. For an ideal gas, if E is the translational kinetic energy of the molecules per unit volume, the pressure is
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Compare P = ⅓ρc² with E = ½ρc².
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Answer: D. P = 2E/3
P = ⅓ρc², and E = ½ρc², so P = ⅔E.
35. Hydrogen and oxygen gases are at the same temperature. Which quantity is the same for their molecules?
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Which kinetic-theory result contains only T?
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Answer: C. Average translational kinetic energy per molecule
Average translational KE per molecule = (3/2)kT, which depends only on temperature. The rms speed and momentum depend on molecular mass, and the mean free path depends on molecular size and number density.
36. The molecules of a gas of density 1.2 kg/m³ have an rms speed of 500 m/s. The pressure of the gas is
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Use the kinetic-theory pressure formula with ⅓.
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Answer: A. 1.0×10⁵ Pa
P = ⅓ρv_rms² = ⅓ × 1.2 × (500)² = 0.4 × 2.5×10⁵ = 1.0×10⁵ Pa.
37. For a diatomic ideal gas whose molecules are rigid (no vibrational modes), the ratio γ = Cp/Cv is
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Count the degrees of freedom: 3 translational + 2 rotational.
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Answer: A. 7/5
There are 5 degrees of freedom, so Cv = 5R/2 and Cp = 7R/2, giving γ = 7/5 = 1.4.
38. One mole of a monatomic ideal gas is mixed with one mole of a rigid diatomic ideal gas. The ratio γ = Cp/Cv of the mixture is
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Average the Cv values weighted by moles, then use Cp = Cv + R.
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Answer: B. 1.50
Cv(mix) = (1 × 3R/2 + 1 × 5R/2)/2 = 2R, and Cp = Cv + R = 3R. So γ = 3R/2R = 1.5. Simply averaging the two γ values (giving 1.53) is wrong.
39. According to Dulong and Petit's law, the molar specific heat of most solid elements at room temperature is about
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Each oscillator contributes both kinetic and potential energy terms.
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Answer: D. 3R
Each atom vibrates in 3 dimensions. Each vibrational mode contributes R (½R kinetic + ½R potential), giving 3R ≈ 25 J mol⁻¹ K⁻¹.
40. If the pressure of a gas is doubled while its temperature is kept constant, the mean free path of its molecules
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Mean free path is inversely proportional to number density.
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Answer: A. becomes half
λ = 1/(√2 πd²n), and n = P/kT. Doubling P at constant T doubles n, so λ is halved.
41. The total internal energy of 2 mol of oxygen (treated as a rigid diatomic ideal gas) at 300 K is (R = 8.3 J mol⁻¹ K⁻¹)
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Apply equipartition with 5 degrees of freedom.
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Answer: B. 12450 J
U = n(f/2)RT with f = 5: U = 2 × 2.5 × 8.3 × 300 = 12450 J. Using f = 3 (translation only) gives 7470 J.
2.5 Laws of thermodynamics
10 questions
42. In an isothermal expansion of an ideal gas,
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What happens to U when T is constant?
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Answer: D. the heat absorbed equals the work done by the gas
For an ideal gas U depends only on T, so ΔU = 0 in an isothermal process. The first law then gives Q = W.
43. A rigid diatomic ideal gas is heated at constant pressure. What fraction of the heat supplied goes into increasing its internal energy?
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Compare Cv with Cp.
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Answer: A. 5/7
ΔU/Q = nCvΔT / nCpΔT = Cv/Cp = (5R/2)/(7R/2) = 5/7. The remaining 2/7 is the work done by the gas.
44. For hydrogen gas (molar mass 2 g/mol), the difference between the principal specific heats, cp − cv, is about (R = 8.3 J mol⁻¹ K⁻¹)
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Convert Mayer's relation from per mole to per kg using M in kg/mol.
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Answer: C. 4150 J kg⁻¹ K⁻¹
Per mole, Cp − Cv = R. Per kilogram, cp − cv = R/M = 8.3/0.002 = 4150 J kg⁻¹ K⁻¹.
45. An ideal monatomic gas (γ = 5/3) is compressed adiabatically to 1/8 of its original volume. Its pressure becomes
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Use PV^γ = constant.
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Answer: D. 32 times
PV^γ = constant, so P₂/P₁ = (V₁/V₂)^γ = 8^(5/3) = (2³)^(5/3) = 2⁵ = 32.
46. A Carnot engine works between a source at 527 °C and a sink at 127 °C. Its efficiency is
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Temperatures must be in kelvin.
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Answer: C. 50%
Convert to kelvin: T₁ = 800 K, T₂ = 400 K. η = 1 − T₂/T₁ = 1 − 400/800 = 0.5. Using Celsius values gives the wrong answer of 75.9%.
47. An ideal (Carnot) refrigerator keeps its freezer at −23 °C in a room at 27 °C. The work needed to remove 1000 J of heat from the freezer is
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Find the coefficient of performance using kelvin temperatures.
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Answer: B. 200 J
COP β = T₂/(T₁ − T₂) = 250/(300 − 250) = 5. W = Q₂/β = 1000/5 = 200 J. The heat rejected to the room would be 1200 J.
48. Which of the following statements is NOT correct?
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Look at the Carnot efficiency formula. What does it depend on?
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Answer: B. The efficiency of a Carnot engine depends on the nature of its working substance
Carnot efficiency, 1 − T₂/T₁, depends only on the reservoir temperatures, not on the working substance. The other three statements are true; the last one is the Kelvin–Planck statement.
49. In which thermodynamic process is all the heat supplied to an ideal gas used to increase its internal energy?
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In which process is the work done zero?
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Answer: A. Isochoric
At constant volume no work is done (W = PΔV = 0), so the first law gives Q = ΔU.
50. An ideal Otto cycle has a compression ratio of 4 and uses a working gas with γ = 1.5. Its efficiency is
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Use η = 1 − 1/r^(γ−1).
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Answer: A. 50%
η = 1 − 1/r^(γ−1) = 1 − 1/4^0.5 = 1 − 1/2 = 50%.
51. 1 kg of ice melts at 0 °C. Taking the latent heat of fusion as 3.36×10⁵ J/kg, the change in entropy of the ice is about
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ΔS = Q/T, with T in kelvin.
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Answer: D. 1.23×10³ J/K
ΔS = Q/T = 3.36×10⁵ / 273 ≈ 1.23×10³ J/K. Entropy increases even though temperature does not change, because heat is absorbed.