Exercises — Thermal Properties of Matter

1. 10.12 marksNCERT Cl-11 Physics Part II, Ch10 Exercises, Q10.1

   The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales.

   Hint. Convert Kelvin to Celsius first with T_C = T_K − 273.15, then convert that Celsius value to Fahrenheit with T_F = (9/5)T_C + 32.

   Step 1 — Convert neon's triple point to Celsius. T_C = T_K − 273.15 = 24.57 − 273.15 = −248.58 °C

   Step 2 — Convert to Fahrenheit. T_F = (9/5)(−248.58) + 32 = −447.44 + 32 = −415.44 °F

   Step 3 — Convert carbon dioxide's triple point to Celsius. T_C = 216.55 − 273.15 = −56.60 °C

   Step 4 — Convert to Fahrenheit. T_F = (9/5)(−56.60) + 32 = −101.88 + 32 = −69.88 °F

   ✦ Neon: −248.58 °C, −415.44 °F. Carbon dioxide: −56.60 °C, −69.88 °F, since both conversions use the same fixed formulas relating the three scales.

   Where students slip. Subtracting 273 instead of 273.15, which shifts every answer by 0.15 °C. The exact conversion always uses 273.15, not the rounded 273.

2. 10.22 marksNCERT Cl-11 Physics Part II, Ch10 Exercises, Q10.2

   Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between T_A and T_B?

   Hint. Both scales are absolute, so they agree that absolute zero is 0 on each. The only thing that differs is the size of one degree, which you can fix using the shared reference point — the triple point of water.

   Step 1 — Recognise both are absolute scales. Both A and B place their zero at absolute zero, so any temperature can be written as a simple multiple of the triple point of water on that scale.

   Step 2 — Write the ratio each scale uses. On scale A, the triple point of water (273.16 K) reads 200 A, so 1 kelvin corresponds to 200/273.16 units of A. Equally, T_A/200 = T(\text{kelvin})/273.16.

   On scale B, the same physical temperature (273.16 K) reads 350 B, so T_B/350 = T(\text{kelvin})/273.16.

   Step 3 — Eliminate the kelvin value between the two. Since both equal the same T(kelvin)/273.16,

   T_A/200 = T_B/350

   Step 4 — Solve for the relation.

   T_A = (200/350) T_B = (4/7) T_B

   ✦ T_A = (4/7) T_B, or equivalently T_B = (7/4) T_A.

   Where students slip. Inverting the ratio to get T_A = (350/200)T_B. The scale with the SMALLER number at the reference point has the LARGER individual degree, so it must take the smaller fraction of the other scale's reading.

3. 10.33 marksNCERT Cl-11 Physics Part II, Ch10 Exercises, Q10.3

   The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law: R = R₀[1 + α(T − T₀)]. The resistance is 101.6 Ω at the triple-point of water 273.16 K, and 165.5 Ω at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Ω?

   Hint. Use the two known points to find α first, then use the same linear law with the unknown resistance to solve for T.

   Step 1 — Use the two known points to find α. At the triple point, R₀ = 101.6 Ω at T₀ = 273.16 K. At the lead point, R = 165.5 Ω at T = 600.5 K.

   Substituting into R = R₀[1 + α(T − T₀)]:

   165.5 = 101.6[1 + α(600.5 − 273.16)] 165.5/101.6 = 1 + α(327.34) 1.6289 − 1 = α(327.34) α = 0.6289/327.34 = 1.9214 × 10⁻³ K⁻¹

   Step 2 — Use the same law for the unknown temperature.

   123.4 = 101.6[1 + α(T − 273.16)] 123.4/101.6 = 1 + α(T − 273.16) 1.2146 − 1 = α(T − 273.16) T − 273.16 = 0.2146/(1.9214 × 10⁻³) = 111.68

   Step 3 — Solve for T.

   T = 273.16 + 111.68 = 384.84 K

   Step 4 — Convert for a sense check. That is about 111.7 °C, a plausible intermediate temperature between the triple point of water (0.01 °C) and the melting point of lead (327.5 °C), which supports the arithmetic.

   ✦ T ≈ 384.8 K (about 111.7 °C).

   Where students slip. Using the two calibration points to set up two separate equations and subtracting incorrectly, losing the R₀ factor. Keep R₀ = 101.6 Ω fixed throughout — it is not an unknown, it is one of the two given calibration values.

4. 10.45 marksNCERT Cl-11 Physics Part II, Ch10 Exercises, Q10.4

   Answer the following: (a) The triple-point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale)? (b) There were two fixed points in the original Celsius scale as mentioned above which were assigned the numbers 0 °C and 100 °C respectively. On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is assigned the number 273.16 K. What is the other fixed point on this (Kelvin) scale? (c) The absolute temperature (Kelvin scale) T is related to the temperature tc on the Celsius scale by tc = T − 273.15. Why do we have 273.15 in this relation, and not 273.16? (d) What is the temperature of the triple-point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale?

   Hint. For (a), think about what makes the melting and boiling points of water unreliable — both depend on something the triple point does not. For (d), use the fact that a Fahrenheit-sized degree is smaller than a kelvin, by the ratio 9/5.

   Step 1 — (a) Why the triple point, not melting and boiling points. The triple point of water is the unique combination of temperature and pressure at which ice, liquid water and water vapour all coexist in equilibrium. It occurs at exactly one pressure, so it needs no external pressure specification.

   The melting point of ice and the boiling point of water, by contrast, both depend on the surrounding pressure — the boiling point especially so, as anyone at high altitude knows. This makes them unreliable as universal fixed points, whereas the triple point is not.

   Step 2 — (b) The other fixed point on the Kelvin scale. The Kelvin scale is an absolute scale, so its other fixed point is absolute zero, 0 K — the point where, by definition, the scale begins. Unlike the original Celsius scale, the Kelvin scale does not need a second calibration point from a physical phenomenon; the zero itself is the second reference.

   Step 3 — (c) Why 273.15 and not 273.16. The number 273.16 is specifically the temperature of the triple point of water. The number 273.15 is the temperature of the ordinary melting point of ice at standard atmospheric pressure — a slightly different physical situation, about 0.01 K lower than the triple point. The relation tc = T − 273.15 is defined so that the Celsius scale's zero coincides with the ordinary melting point of ice, which is the everyday reference point, not the triple point used to fix the Kelvin scale itself.

   Step 4 — (d) Triple point on a Fahrenheit-sized absolute scale. A Fahrenheit degree is smaller than a kelvin by the factor 5/9, so a scale using Fahrenheit-sized degrees needs more of them to span the same physical temperature — a factor of 9/5 more units.

   The triple point is 273.16 K above absolute zero. On the new scale:

   T = 273.16 × (9/5) = 491.69

   ✦ (a) the triple point needs no pressure specification, unlike melting/boiling points; (b) absolute zero, 0 K; (c) 273.15 is the ordinary ice point, a separate reference from the triple point's 273.16; (d) 491.69 units.

   Where students slip. Answering (d) with 273.16 × (5/9), inverting the ratio. Fahrenheit degrees are smaller, so more of them are needed to cover the same span — the multiplying factor must be greater than 1, which only 9/5 gives.

5. 10.55 marksNCERT Cl-11 Physics Part II, Ch10 Exercises, Q10.5

   Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made: Triple-point of water — thermometer A: 1.250×10⁵ Pa, thermometer B: 0.200×10⁵ Pa. Normal melting point of sulphur — thermometer A: 1.797×10⁵ Pa, thermometer B: 0.287×10⁵ Pa. (a) What is the absolute temperature of normal melting point of sulphur as read by thermometers A and B? (b) What do you think is the reason behind the slight difference in answers of thermometers A and B? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings?

   Hint. A constant-volume gas thermometer gives T ∝ p, calibrated against the triple point of water at 273.16 K. For (b), think about what an 'ideal' gas assumption leaves out at the pressures actually used.

   Step 1 — (a) Use T ∝ p with the triple point as calibration. For thermometer A:

   T_A = 273.16 × (p_S/p_triple) = 273.16 × (1.797×10⁵/1.250×10⁵) = 273.16 × 1.4376 = 392.69 K

   For thermometer B:

   T_B = 273.16 × (0.287×10⁵/0.200×10⁵) = 273.16 × 1.435 = 391.98 K

   Step 2 — (b) Why the two disagree, though neither is faulty. The ideal gas law, on which a constant-volume gas thermometer's temperature scale rests, is only exactly true in the limit of very low pressure, where the gas molecules are far enough apart that intermolecular forces and the volume of the molecules themselves become negligible.

   At the pressures actually used here, oxygen and hydrogen depart from ideal behaviour by slightly different amounts — real gases are never perfectly ideal, and different gases deviate differently. That is the source of the small disagreement between A and B, even though both instruments are working correctly.

   Step 3 — What reduces the discrepancy. Repeat the experiment using successively lower and lower gas pressure in the bulb of each thermometer, and extrapolate the resulting readings to the limit of zero pressure. As pressure falls, both gases approach ideal behaviour, and their extrapolated readings converge to the same value.

   ✦ (a) T_A ≈ 392.69 K, T_B ≈ 391.98 K. (b) real gases deviate from ideal behaviour at finite pressure, and differently for different gases; extrapolating readings to zero pressure removes the discrepancy.

   Where students slip. Attributing the difference to one thermometer being miscalibrated. The question states neither is faulty — the disagreement is a real physical limitation of using any real gas at finite pressure to define temperature, not an instrument fault.

6. 10.65 marksNCERT Cl-11 Physics Part II, Ch10 Exercises, Q10.6

   A steel tape 1m long is correctly calibrated for a temperature of 27.0 °C. The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0 °C. What is the actual length of the steel rod on that day? What is the length of the same steel rod on a day when the temperature is 27.0 °C? Coefficient of linear expansion of steel = 1.20 × 10⁻⁵ K⁻¹.

   Hint. At 45°C the tape's own markings have stretched, so a reading of 63.0 cm on the tape corresponds to a slightly larger true length. Once you have the true length at 45°C, cool the rod itself back down to 27°C using the same expansion coefficient.

   Step 1 — Find the actual length of the rod at 45°C. The tape is correct at 27°C, but at 45°C it has expanded, so each marked centimetre is now slightly more than a true centimetre. A reading of 63.0 marked-cm therefore corresponds to a true length of

   L(45°C) = 63.0 × [1 + α(45.0 − 27.0)] = 63.0 × [1 + (1.20×10⁻⁵)(18.0)] = 63.0 × 1.000216 = 63.0136 cm

   Step 2 — Find the length of the same rod at 27°C. The rod itself is made of steel too, so cooling it from 45°C back to 27°C shrinks it by the same expansion law:

   L(27°C) = L(45°C)/[1 + α(45.0 − 27.0)] = 63.0136/1.000216 ≈ 63.0000 cm

   Step 3 — Notice why the two nearly cancel. Because the tape and the rod are both steel with the same expansion coefficient, the tape's own distortion at 45°C very nearly compensates for the rod's own expansion — a tape made of the same material as what it measures reads the object's length at the calibration temperature almost exactly, regardless of the temperature at the moment of measurement.

   ✦ Actual length at 45°C ≈ 63.0136 cm; length of the same rod at 27°C ≈ 63.0000 cm.

   Where students slip. Treating the 63.0 cm reading as already the true length and only converting once. The reading is in tape-units, not true centimetres, so it must first be corrected for the tape's own expansion before anything else is done with it.

7. 10.73 marksNCERT Cl-11 Physics Part II, Ch10 Exercises, Q10.7

   A large steel wheel is to be fitted on to a shaft of the same material. At 27 °C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. The shaft is cooled using 'dry ice'. At what temperature of the shaft does the wheel slip on the shaft? Assume coefficient of linear expansion of the steel to be constant over the required temperature range: α_steel = 1.20 × 10⁻⁵ K⁻¹.

   Hint. The shaft must shrink until its diameter equals the (fixed) diameter of the wheel's hole. Set up d(T) = d₀[1 + α(T − T₀)] and solve for the temperature that gives the smaller diameter.

   Step 1 — Write the shaft's diameter as a function of temperature.

   d(T) = d₀[1 + α(T − T₀)]

   with d₀ = 8.70 cm at T₀ = 27°C.

   Step 2 — Set the shaft's diameter equal to the hole's diameter. The wheel slips onto the shaft exactly when the shaft has shrunk to 8.69 cm:

   8.69 = 8.70[1 + α(T − 27)] 8.69/8.70 = 1 + α(T − 27) 0.99885 − 1 = α(T − 27) −1.1494×10⁻³ = (1.20×10⁻⁵)(T − 27)

   Step 3 — Solve for T.

   T − 27 = −1.1494×10⁻³/1.20×10⁻⁵ = −95.79 T = 27 − 95.79 = −68.79°C

   Step 4 — Check this is a reasonable temperature for dry ice. Dry ice (solid carbon dioxide) sublimes at about −78.5°C, so a shaft cooled by dry ice can certainly reach −68.79°C. The numbers are consistent with the physical method described.

   ✦ The wheel slips onto the shaft at about −68.8°C.

   Where students slip. Using the wheel's hole diameter as the one that changes with temperature instead of the shaft's. It is the shaft that is cooled and must shrink; the hole in the wheel is fixed at 8.69 cm throughout.

8. 10.82 marksNCERT Cl-11 Physics Part II, Ch10 Exercises, Q10.8

   A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0 °C. What is the change in the diameter of the hole when the sheet is heated to 227 °C? Coefficient of linear expansion of copper = 1.70 × 10⁻⁵ K⁻¹.

   Hint. A hole in a sheet expands exactly as though it were a solid disc of the same material — this is a standard but easy-to-doubt fact worth stating explicitly.

   Step 1 — Recall how a hole behaves under thermal expansion. A hole in a solid sheet expands as though it were filled with the same material as the sheet — heating the sheet makes the hole grow, exactly like linear expansion of a solid disc of that diameter. This is not intuitive but is a standard and examinable result.

   Step 2 — Apply the linear expansion formula to the diameter.

   Δd = d₀ α ΔT = 4.24 × (1.70×10⁻⁵) × (227 − 27) = 4.24 × 1.70×10⁻⁵ × 200 = 0.01442 cm

   Step 3 — State the result and its sign. The change is positive — the hole gets larger, not smaller, since it expands along with the surrounding metal rather than closing in.

   ✦ Δd ≈ 0.0144 cm — the hole enlarges by about 0.014 cm.

   Where students slip. Assuming the hole shrinks because the surrounding material is expanding "inward" toward the hole. A hole expands outward together with the material around it, growing exactly as a solid piece of that diameter would.

9. 10.93 marksNCERT Cl-11 Physics Part II, Ch10 Exercises, Q10.9

   A brass wire 1.8 m long at 27 °C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of −39 °C, what is the tension developed in the wire, if its diameter is 2.0 mm? Coefficient of linear expansion of brass = 2.0 × 10⁻⁵ K⁻¹; Young's modulus of brass = 0.91 × 10¹¹ Pa.

   Hint. The wire wants to contract as it cools but the rigid supports will not let it, so the whole thermal contraction shows up as an elastic strain instead. Use that strain with Young's modulus to find the stress, then the force.

   Step 1 — Find the fractional contraction the wire would undergo if free.

   ΔL/L = α ΔT = (2.0×10⁻⁵)(39 − (−(−39)))

   More carefully: ΔT = 27 − (−39) = 66°C, so

   ΔL/L = α ΔT = (2.0×10⁻⁵)(66) = 1.32×10⁻³

   Step 2 — Recognise the rigid supports prevent this contraction. Since the wire cannot actually shorten, the would-be contraction is instead entirely converted into a tensile strain. So the strain in Young's modulus is exactly this value:

   strain = ΔL/L = 1.32×10⁻³

   Step 3 — Find the stress from Young's modulus.

   stress = Y × strain = (0.91×10¹¹)(1.32×10⁻³) = 1.201×10⁸ Pa

   Step 4 — Convert stress to tension using the wire's area.

   A = π(d/2)² = π(1.0×10⁻³)² = 3.1416×10⁻⁶ m²

   T = stress × A = (1.201×10⁸)(3.1416×10⁻⁶) = 377.4 N

   ✦ Tension developed ≈ 377 N.

   Where students slip. Using the diameter directly in the area formula instead of the radius. The area needs (d/2)², and using d² alone would overstate the area — and hence the tension — by a factor of four.

10. 10.105 marksNCERT Cl-11 Physics Part II, Ch10 Exercises, Q10.10

    A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250 °C, if the original lengths are at 40.0 °C? Is there a 'thermal stress' developed at the junction? The ends of the rod are free to expand (Coefficient of linear expansion of brass = 2.0 × 10⁻⁵ K⁻¹, steel = 1.2 × 10⁻⁵ K⁻¹).

    Hint. Each rod expands independently by its own coefficient, since the ends are free — add the two changes for the total. For the thermal stress question, ask what actually causes a stress: is it different expansion coefficients on their own, or is it something constraining that expansion?

    Step 1 — Find the change in length of each rod separately. ΔT = 250 − 40.0 = 210°C, L = 0.50 m for each rod.

    Brass: ΔL_brass = L α_brass ΔT = 0.50 × (2.0×10⁻⁵) × 210 = 2.10×10⁻³ m = 0.210 cm

    Steel: ΔL_steel = L α_steel ΔT = 0.50 × (1.2×10⁻⁵) × 210 = 1.26×10⁻³ m = 0.126 cm

    Step 2 — Add the two for the total change.

    ΔL_total = 0.210 + 0.126 = 0.336 cm

    Step 3 — Decide whether there is a thermal stress. A thermal stress arises only when a rod is prevented from expanding or contracting freely and the resulting strain shows up as an elastic stress instead — exactly the situation in the previous question, where rigid supports blocked the contraction.

    Here, the question states explicitly that the ends of the rod are free to expand. Each material is left free to change length by its own natural amount, with nothing constraining it. Since no constraint prevents the expansion, no stress is developed anywhere, including at the junction.

    ✦ Total change in length ≈ 0.336 cm; no thermal stress develops, because both ends are free to expand.

    Where students slip. Assuming different expansion coefficients at a junction automatically create a stress. A stress requires a constraint against expansion — different coefficients alone, with both ends free, only produce different but unimpeded elongations, and no internal force.

11. 10.112 marksNCERT Cl-11 Physics Part II, Ch10 Exercises, Q10.11

    The coefficient of volume expansion of glycerine is 49 × 10⁻⁵ K⁻¹. What is the fractional change in its density for a 30 °C rise in temperature?

    Hint. Mass stays fixed while volume changes, so density and volume are inversely related. For a small fractional change, the fractional decrease in density is very nearly equal in size to the fractional increase in volume.

    Step 1 — Find the fractional change in volume.

    ΔV/V = γ ΔT = (49×10⁻⁵)(30) = 1.47×10⁻² = 0.0147

    Step 2 — Relate this to the fractional change in density. Since ρ = m/V and the mass m stays constant as the liquid expands, ρV = constant, so a small fractional increase in volume corresponds to an equal and opposite small fractional decrease in density:

    Δρ/ρ ≈ −ΔV/V = −1.47×10⁻²

    Step 3 — State the result. The density decreases by about 1.47%, which is the same size as the volume's fractional increase, just with the opposite sign.

    ✦ Fractional change in density ≈ −1.47 × 10⁻² (a decrease of about 1.47%).

    Where students slip. Reporting the answer as a positive number, or as an increase. Heating always expands a liquid's volume and correspondingly lowers its density, so the fractional change in density must come out negative.

12. 10.123 marksNCERT Cl-11 Physics Part II, Ch10 Exercises, Q10.12

    A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings. Specific heat of aluminium = 0.91 J g⁻¹ K⁻¹.

    Hint. Only half the power actually goes into the aluminium block. Convert the specific heat to J/kg K before using it with a mass in kilograms.

    Step 1 — Find the total energy supplied.

    E_total = P × t = (10×10³ W) × (2.5 × 60 s) = 10,000 × 150 = 1.5×10⁶ J

    Step 2 — Find the energy that actually heats the block. Only 50% goes into the block; the rest heats the machine or is lost:

    Q = 0.50 × 1.5×10⁶ = 7.5×10⁵ J

    Step 3 — Convert the specific heat to SI units.

    s = 0.91 J g⁻¹ K⁻¹ = 0.91×10³ J kg⁻¹ K⁻¹ = 910 J kg⁻¹ K⁻¹

    Step 4 — Find the temperature rise.

    ΔT = Q/(ms) = (7.5×10⁵)/(8.0 × 910) = (7.5×10⁵)/7280 ≈ 103.0°C, which is why the factor of 0.50 must be applied before dividing, since only that portion actually raises the block's temperature.

    ✦ Temperature rise ≈ 103 °C.

    Where students slip. Forgetting the factor of 0.50 and using the full power delivered. Only half the energy supplied by the machine actually ends up heating the aluminium block; the rest is explicitly stated to go elsewhere.

13. 10.133 marksNCERT Cl-11 Physics Part II, Ch10 Exercises, Q10.13

    A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500 °C and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper = 0.39 J g⁻¹ K⁻¹; heat of fusion of water = 335 J g⁻¹).

    Hint. The final temperature of the copper, once it stops giving up heat, is 0°C — the temperature of the ice it is resting on. All the heat the copper gives up in cooling from 500°C to 0°C goes into melting ice.

    Step 1 — Find the heat released by the copper as it cools. The copper cools from 500°C down to 0°C, the temperature of the ice (assuming the ice block is large enough to stay at 0°C throughout):

    Q = m_Cu × s_Cu × ΔT

    Using SI units: m_Cu = 2.5 kg, s_Cu = 0.39×10³ J kg⁻¹ K⁻¹, ΔT = 500°C

    Q = 2.5 × 390 × 500 = 487,500 J

    Step 2 — Convert this heat into a mass of ice melted. All of this heat goes into melting ice at its fusion temperature, so

    m_ice = Q/L_f = 487,500/(335×10³) = 1.455 kg

    Step 3 — Note why this is the maximum amount. This assumes every joule the copper releases goes into melting ice, with none lost to the surroundings — which is exactly why the question asks for the maximum possible amount, an upper bound rather than a guaranteed outcome.

    ✦ Maximum ice melted ≈ 1.46 kg.

    Where students slip. Using specific heat units of J g⁻¹ K⁻¹ directly with a mass in kilograms without converting. Either convert the specific heat to J kg⁻¹ K⁻¹ (multiply by 1000) or convert the mass to grams — mixing the two unit systems gives an answer a thousand times too small or too large.

14. 10.145 marksNCERT Cl-11 Physics Part II, Ch10 Exercises, Q10.14

    In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150 °C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm³ of water at 27 °C. The final temperature is 40 °C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal?

    Hint. Use heat lost by the metal equals heat gained by the water and the calorimeter (represented by its water equivalent). For the second part, think about where the extra heat goes if losses to the surroundings are not actually zero.

    Step 1 — Find the mass of water. 150 cm³ of water has mass 0.150 kg (density 1000 kg/m³).

    Step 2 — Write the heat balance. Heat lost by the metal = heat gained by the water + heat gained by the calorimeter (via its water equivalent):

    m_metal s_metal (150 − 40) = (m_water + W) s_water (40 − 27)

    Step 3 — Substitute the known values.

    0.20 × s_metal × 110 = (0.150 + 0.025) × 4186 × 13 0.20 × s_metal × 110 = 0.175 × 4186 × 13 22 s_metal = 9523.2 s_metal = 432.9 J kg⁻¹ K⁻¹

    Step 4 — Answer the second part. If heat losses to the surroundings are not negligible, then some of the heat the metal actually released escaped to the surroundings rather than going into the water and calorimeter. The calculation above only accounts for the heat that reached the water and calorimeter, which is less than the true total heat released by the metal.

    Since the calculated specific heat comes from dividing this understated heat by the metal's actual temperature drop, the calculated value of s_metal will be smaller than the actual specific heat of the metal.

    ✦ Calculated specific heat ≈ 433 J kg⁻¹ K⁻¹; this is smaller than the true value, because unaccounted heat loss to the surroundings means the metal actually released more heat than the water and calorimeter show.

    Where students slip. Forgetting to include the calorimeter's water equivalent alongside the water itself. The water equivalent represents the calorimeter's own heat capacity expressed as an equivalent mass of water, and omitting it understates the heat absorbed.

15. 10.155 marksNCERT Cl-11 Physics Part II, Ch10 Exercises, Q10.15 — the standard explanation draws on degrees of freedom, which belongs to Chapter 12 (Kinetic Theory), not this chapter

    Given below are observations on molar specific heats at room temperature of some common gases: Hydrogen 4.87, Nitrogen 4.97, Oxygen 5.02, Nitric oxide 4.99, Carbon monoxide 5.01, Chlorine 6.17 (all in cal mol⁻¹ K⁻¹). The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, molar specific heat of a monatomic gas is 2.92 cal/mol K. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine?

    Hint. This chapter never derives molar specific heats from molecular structure — that reasoning belongs to the kinetic theory of gases, two chapters ahead. The explanation is about how many independent ways a molecule can store energy.

    Step 1 — Note where this explanation actually comes from. This chapter presents molar specific heats as measured quantities but never derives their values from molecular structure. The full explanation properly belongs to the kinetic theory of gases (Chapter 12), which introduces the idea of degrees of freedom — here it is, since the exercise expects it.

    Step 2 — Why monatomic gases have the smallest value. A monatomic gas molecule, being a single atom, can only store energy in its translational motion — moving along three independent directions in space. By the equipartition of energy, each of these three degrees of freedom contributes (1/2)R to the molar specific heat at constant volume, giving Cv = (3/2)R ≈ 2.98 cal/mol K, matching the quoted 2.92 cal/mol K closely.

    Step 3 — Why the listed gases have a larger value. Hydrogen, nitrogen, oxygen, nitric oxide and carbon monoxide are all diatomic gases. In addition to the three translational degrees of freedom, a diatomic molecule can also rotate about two independent axes perpendicular to the bond, adding two more degrees of freedom. This gives Cv = (5/2)R ≈ 4.97 cal/mol K — very close to the values listed for nitrogen, oxygen, nitric oxide and carbon monoxide.

    Step 4 — What the larger value for chlorine suggests. Chlorine's value, 6.17 cal/mol K, is noticeably larger than the (5/2)R prediction of about 4.97. This suggests that at room temperature, chlorine's vibrational degree of freedom — the two atoms oscillating along the bond — is already becoming thermally active and contributing extra energy storage, on top of the translational and rotational contributions the other diatomic gases show. This is consistent with a heavier or more weakly bonded diatomic molecule having a lower vibrational frequency, so vibration becomes significant at a lower temperature than for lighter molecules like hydrogen or nitrogen.

    ✦ Monatomic gases store energy only in translation, giving Cv = (3/2)R. Diatomic gases add two rotational degrees of freedom, giving Cv = (5/2)R, matching most of the listed values. Chlorine's higher value suggests its vibrational mode is already contributing at room temperature.

    Where students slip. Trying to explain the difference using only the material in this chapter (specific heat capacity, calorimetry). The reasoning genuinely requires the kinetic theory of gases and the concept of degrees of freedom, introduced later in the book.

16. 10.165 marksNCERT Cl-11 Physics Part II, Ch10 Exercises, Q10.16

    A child running a temperature of 101°F is given an antipyrin (i.e. a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98 °F in 20 minutes, what is the average rate of extra evaporation caused, by the drug. Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of human body is approximately the same as that of water, and latent heat of evaporation of water at that temperature is about 580 cal g⁻¹.

    Hint. Convert the Fahrenheit temperature drop to Celsius or kelvin before using it in Q = ms ΔT. The evaporated sweat is what carries the heat away, using the latent heat of evaporation, not the specific heat.

    Step 1 — Convert the temperature drop to Celsius. ΔT_F = 101 − 98 = 3°F. Since a Fahrenheit degree is 5/9 the size of a Celsius degree,

    ΔT_C = 3 × (5/9) = 1.667°C

    Step 2 — Find the heat that must be removed from the child's body. Using the specific heat of water for the body:

    Q = m s ΔT = 30 × 4186 × 1.667 = 2.093×10⁵ J

    Step 3 — Convert the latent heat of evaporation to SI units.

    L = 580 cal/g = 580 × 4.186 J/g = 2427.9 J/g = 2.4279×10⁶ J/kg

    Step 4 — Find the mass of sweat evaporated.

    m_evap = Q/L = (2.093×10⁵)/(2.4279×10⁶) = 0.08621 kg = 86.21 g

    Step 5 — Convert to a rate over the 20 minutes.

    rate = 86.21 g / 20 min ≈ 4.31 g/min (about 0.072 g/s)

    ✦ Average extra rate of evaporation ≈ 4.3 g per minute.

    Where students slip. Using the specific heat of water to find how much sweat is needed, rather than the latent heat of evaporation. Specific heat governs a temperature change of the sweat itself; here the sweat carries heat away by changing phase, from liquid to vapour, which needs the latent heat.

17. 10.175 marksNCERT Cl-11 Physics Part II, Ch10 Exercises, Q10.17

    A 'thermacole' icebox is a cheap and an efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6 h. The outside temperature is 45 °C, and coefficient of thermal conductivity of thermacole is 0.01 J s⁻¹ m⁻¹ K⁻¹. [Heat of fusion of water = 335 × 10³ J kg⁻¹]

    Hint. Find the heat conducted into the box per second through all six faces combined, then multiply by the total time to get the total heat that enters over 6 hours. That heat melts ice at a constant 0°C inside.

    Step 1 — Find the total surface area of the cube. Side = 0.30 m, so one face has area (0.30)² = 0.09 m². A cube has 6 faces:

    A = 6 × 0.09 = 0.54 m²

    Step 2 — Find the rate of heat conduction into the box. Using the conduction formula, with the ice inside at 0°C and outside at 45°C, so ΔT = 45 K:

    H = KA ΔT / d = (0.01)(0.54)(45)/(0.05) = 0.243/0.05 = 4.86 J/s

    Step 3 — Find the total heat entering over 6 hours.

    t = 6 × 3600 = 21,600 s

    Q = H × t = 4.86 × 21,600 = 1.05×10⁵ J

    Step 4 — Find the mass of ice melted by this heat.

    m_melted = Q/L_f = (1.05×10⁵)/(335×10³) = 0.313 kg

    Step 5 — Find the ice remaining.

    m_remaining = 4.0 − 0.313 = 3.687 kg

    ✦ About 3.69 kg of ice remains after 6 hours — the icebox loses roughly 0.31 kg of ice to melting.

    Where students slip. Using the area of a single face rather than all six. Heat enters through every face of the cube simultaneously, so the conduction formula needs the total surface area, not just one side.

18. 10.183 marksNCERT Cl-11 Physics Part II, Ch10 Exercises, Q10.18

    A brass boiler has a base area of 0.15 m² and thickness 1.0 cm. It boils water at the rate of 6.0 kg/min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = 109 J s⁻¹ m⁻¹ K⁻¹; Heat of vaporisation of water = 2256 × 10³ J kg⁻¹.

    Hint. The heat conducted through the base per second must equal the heat needed to vaporise water at that same rate. Set the conduction formula equal to that rate and solve for the temperature difference across the base.

    Step 1 — Find the rate at which heat must be supplied. Water boils away at 6.0 kg/min = 0.10 kg/s. Vaporising it needs:

    H = (dm/dt) × L_v = 0.10 × 2256×10³ = 2.256×10⁵ J/s

    Step 2 — Use the conduction formula to find the temperature difference across the base.

    H = KA ΔT/d ⟹ ΔT = Hd/(KA)

    ΔT = (2.256×10⁵)(0.010)/[(109)(0.15)] = 2256/16.35 = 137.98 K

    Step 3 — Add the boiling point of water to get the flame-side temperature. The inner surface of the base is at 100°C (boiling water). The outer, flame-facing surface must be hotter by ΔT:

    T_flame = 100 + 137.98 = 238.0°C, since the base must sit that much hotter on the flame side to drive the required heat flow through to the boiling water.

    ✦ Temperature of the flame in contact with the boiler ≈ 238 °C.

    Where students slip. Reporting ΔT itself as the final answer. ΔT is only the temperature difference across the base — the flame-side temperature is 100°C higher still, since the inner surface stays at the boiling point of water.

19. 10.195 marksNCERT Cl-11 Physics Part II, Ch10 Exercises, Q10.19 — part (a) needs Kirchhoff's radiation law, which does not appear in the 2026-27 chapter

    Explain why: (a) a body with large reflectivity is a poor emitter (b) a brass tumbler feels much colder than a wooden tray on a chilly day (c) an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace (d) the earth without its atmosphere would be inhospitably cold (e) heating systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water

    Hint. For (a), reflectivity and absorptivity are complementary for an opaque body — a highly reflective surface absorbs very little. Kirchhoff's law of radiation, which links good absorption to good emission, is not stated in this chapter but is needed here. For (b) compare the thermal conductivities from the chapter's own table. For (e) compare the extra latent heat steam carries.

    Step 1 — (a) Large reflectivity means a poor emitter. This part relies on Kirchhoff's law of radiation — a good absorber of radiation is also a good emitter of it — which is not stated anywhere in the 2026-27 chapter, though the exercise needs it.

    A body with large reflectivity, by definition, reflects most of the radiation falling on it and absorbs very little — it is a poor absorber. By Kirchhoff's law, a poor absorber is equally a poor emitter, since emission and absorption are two sides of the same equilibrium process at a given wavelength. Hence a highly reflective body, like a polished metal surface, radiates heat away only slowly compared with a dull, dark surface at the same temperature.

    Step 2 — (b) Brass feels colder than wood. From the chapter's own table of thermal conductivities, brass conducts heat at about 109 J s⁻¹m⁻¹K⁻¹, while wood conducts at only about 0.12 — brass is nearly a thousand times better a conductor. Touching brass on a cold day draws heat away from your hand very rapidly, because brass conducts that heat inward efficiently, while wood, a poor conductor, barely draws any heat from your skin at all. Both objects are at the same room or outdoor temperature; what differs is how fast each conducts heat away from your hand.

    Step 3 — (c) The optical pyrometer discrepancy. An optical pyrometer calibrated for an ideal black body assumes the object being measured absorbs and emits perfectly (emissivity e = 1). A red-hot iron piece in the open has an emissivity less than 1, so it actually emits less radiation at a given true temperature than a perfect black body would — the pyrometer, expecting more radiation for that temperature, underestimates it.

    Inside a furnace, the iron piece is surrounded by other hot walls also radiating strongly, and repeated reflection and re-absorption inside the enclosed cavity makes the cavity behave as a near-perfect black body radiator regardless of the iron's own emissivity. In that setting the pyrometer reads correctly.

    Step 4 — (d) Earth without its atmosphere. The atmosphere absorbs and re-radiates a significant portion of the infrared radiation the Earth's surface emits back toward space, warming the surface — the mechanism behind the natural greenhouse effect. Strip away the atmosphere and that outgoing radiation escapes to space unimpeded, so the surface would settle to a much lower equilibrium temperature, making the Earth inhospitably cold.

    Step 5 — (e) Steam heating is more efficient than hot water. Steam at 100°C carries not only the same sensible heat that hot water at 100°C would when cooling down, but also releases its large latent heat of vaporisation, about 2256×10³ J/kg, as it condenses back into water inside the radiator. This extra latent heat, delivered at essentially constant temperature, makes a steam-based system deliver considerably more heat per kilogram of fluid circulated than a hot-water system that only cools through a modest temperature drop.

    ✦ (a) poor absorbers are poor emitters, by Kirchhoff's law; (b) brass conducts heat away from the hand far faster than wood; (c) emissivity below 1 in the open, but furnace behaves as an ideal cavity; (d) the atmosphere traps outgoing radiation; (e) condensing steam releases a large additional latent heat.

    Where students slip. Explaining (b) by claiming brass is "naturally colder" than wood. Both are at the same ambient temperature; the sensation of cold is entirely about the rate of heat conduction away from the skin, not about any difference in the objects' actual temperatures.

20. 10.205 marksNCERT Cl-11 Physics Part II, Ch10 Exercises, Q10.20

    A body cools from 80 °C to 50 °C in 5 minutes. Calculate the time it takes to cool from 60 °C to 30 °C. The temperature of the surroundings is 20 °C.

    Hint. Newton's law of cooling gives an exponential decay of (T − T_surroundings) with time. Use the first cooling interval to find the decay constant, then apply the same constant to the second interval.

    Step 1 — Write Newton's law of cooling in its exponential form. For a body cooling toward surroundings at T_s, the excess temperature (T − T_s) decays exponentially:

    T − T_s = (T_initial − T_s) e^{−kt}

    Step 2 — Use the first interval to find k. Here T_s = 20°C. From 80°C to 50°C in 5 minutes:

    (50 − 20) = (80 − 20) e^{−k(5)} 30/60 = e^{−5k} ln(0.5) = −5k k = ln(2)/5 = 0.6931/5 = 0.1386 min⁻¹

    Step 3 — Apply the same k to the second interval. From 60°C to 30°C, again with T_s = 20°C:

    (30 − 20) = (60 − 20) e^{−kt} 10/40 = e^{−kt} ln(0.25) = −kt t = ln(4)/k = 1.3863/0.1386 = 10.0 min

    Step 4 — Sanity check using the shape of the decay. Going from 60°C to 30°C is a drop of 40°C in excess temperature down to 10°C — a factor of 4 — compared with the first interval's factor of 2 (60°C excess down to 30°C excess). Since the decay is exponential, halving twice (a factor of 4) should take exactly twice as long as halving once, and indeed 10 minutes is exactly double the original 5 minutes.

    ✦ Time to cool from 60°C to 30°C ≈ 10 minutes.

    Where students slip. Using the simplified linear form of Newton's law (rate proportional to the excess temperature at a single instant, applied crudely as an average) without solving the exponential equation properly. Both intervals here are exact multiples of the same decay factor, so the exponential method gives the clean, correct answer of exactly 10 minutes.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 11 Physics Part II textbook, Reprint 2026-27 (keph203.pdf, 24 pages), cross-checked against the official CBSE curriculum 2026-27, Subject Code 042, Unit VII — one end-of-chapter Exercises set (20 questions, 10.1-10.20). Two orphaned-exercise findings: Kirchhoff's radiation law (good absorbers are good emitters) is never stated in the 2026-27 chapter yet Q10.19(a) depends on it, and the degrees-of-freedom explanation Q10.15 asks for belongs to Chapter 12 (Kinetic Theory), two chapters later.. Questions are referenced from the NCERT textbook for identification.