NCERT Solutions for Class 11 Physics Chapter 10: Thermal Properties of Matter

Concept. The Kelvin and Celsius scales have the same size of degree; they differ only in the zero: t_C = T − 273.15. The Fahrenheit scale is related to Celsius by t_F = (9/5)t_C + 32. So every Kelvin value is first moved to Celsius and then, if needed, to Fahrenheit.

Question. 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.

Given

• Triple point of neon, T₁ = 24.57 K
• Triple point of carbon dioxide, T₂ = 216.55 K
• Conversion: t_C = T − 273.15 and t_F = (9/5)t_C + 32

Solution: neon

t_C = 24.57 − 273.15 = −248.58 °C

t_F = (9/5)(−248.58) + 32 = −447.444 + 32 = −415.44 °F

Solution: carbon dioxide

t_C = 216.55 − 273.15 = −56.60 °C

t_F = (9/5)(−56.60) + 32 = −101.88 + 32 = −69.88 °F

Result

Why the order of conversion matters. A temperature difference of 1 K equals 1 °C but equals 9/5 = 1.8 °F. The factor 32 appears only when we convert an actual temperature reading (a point on the scale) and never when we convert a difference. Many students wrongly add 32 while converting a rise of temperature; remember that a rise of 10 °C is a rise of 18 °F, not 50 °F.

Sanity check. Carbon dioxide at 216.55 K is below the freezing point of water (273.15 K), so a negative Celsius value is expected. Neon at only 24.57 K is extremely close to absolute zero (−273.15 °C), so its Celsius value (−248.58 °C) must be only about 24.6 degrees above −273.15 — and it is. The Fahrenheit absolute zero is −459.67 °F, and −415.44 °F is about 44.2 °F above it, which is 24.57 × 1.8 = 44.2 °F. The check works.

Extra depth. Neon and carbon dioxide triple points are also used as fixed points for low-temperature thermometry: the International Temperature Scale of 1990 (ITS-90) lists the triple points of hydrogen (13.8033 K), neon (24.5561 K), oxygen (54.3584 K), argon (83.8058 K) and water (273.16 K). The triple point is chosen because at that single temperature and pressure the solid, liquid and gas phases coexist, so the temperature is perfectly reproducible in any laboratory in the world.

Conclusion: Neon: −248.58 °C = −415.44 °F; carbon dioxide: −56.60 °C = −69.88 °F.

Concept. An absolute scale has its zero at absolute zero, so the reading is directly proportional to the thermodynamic temperature. Two such scales therefore differ only by a constant factor, fixed by assigning different numbers to the same fixed point (the triple point of water).

Question. 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?

Given

• Triple point of water on scale A = 200 A
• Triple point of water on scale B = 350 B
• Both scales are absolute, so zero on each is absolute zero

Solution

On an absolute scale T ∝ (thermodynamic temperature). Hence for the same physical temperature the readings on A and B have a fixed ratio, equal to the ratio of their triple-point values:

T_A / 200 = T_B / 350

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

Check with the Kelvin scale

The Kelvin scale gives the triple point 273.16 K. Then T_A = (200/273.16)T = 0.7322 T and T_B = (350/273.16)T = 1.2813 T, where T is the Kelvin value. Dividing, T_A/T_B = 0.7322/1.2813 = 0.5714 = 4/7, which confirms the result.

Why no constant term appears

If a scale had a zero other than absolute zero (like Celsius or Fahrenheit) the relation would be linear with an intercept, e.g. t_F = (9/5)t_C + 32. For two absolute scales, absolute zero is 0 on both, so the straight line passes through the origin and only a proportionality factor remains. This is exactly why the problem can be solved from a single fixed point.

Interpretation. The size of one degree on scale A is 273.16/200 = 1.366 K, whereas one degree on scale B is 273.16/350 = 0.780 K; hence scale B has the finer degree. A body at 350 B (= 273.16 K = 0.01 °C) reads 200 A, as required.

Extra depth. Real-life absolute scales are Kelvin (SI unit; 1 K = 1/273.16 of the thermodynamic temperature of the triple point of water) and Rankine (T_R = 1.8 T_K, degree size equal to the Fahrenheit degree). In the 2019 SI revision the kelvin is now defined by fixing the Boltzmann constant k = 1.380649 × 10⁻²³ J K⁻¹, but the triple point of water is still 273.16 K to within experimental error.

Conclusion: T_A = (4/7) T_B.

Concept. For a resistance thermometer, R = R₀[1 + α(T − T₀)]. Two known calibration points fix the two unknown constants (R₀ is the resistance at T₀, α is the temperature coefficient). After finding α we invert the relation for the unknown temperature.

Question. The resistance in ohms of a thermometer varies with temperature as 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?

Given

• T₀ = 273.16 K with R₀ = 101.6 Ω
• At T₁ = 600.5 K, R₁ = 165.5 Ω
• Unknown: T when R = 123.4 Ω

Step 1: find α

165.5 = 101.6[1 + α(600.5 − 273.16)]

165.5/101.6 = 1.6289 = 1 + α(327.34)

α = 0.6289/327.34 = 1.921 × 10⁻³ K⁻¹

Step 2: find T at R = 123.4 Ω

123.4/101.6 = 1.2146 = 1 + α(T − 273.16)

T − 273.16 = 0.2146/(1.921 × 10⁻³) = 111.7 K

T = 273.16 + 111.7 = 384.8 K

Shortcut that avoids α

Because R is linear in T, the ratio method is faster: (R − R₀)/(R₁ − R₀) = (T − T₀)/(T₁ − T₀). Here (123.4 − 101.6)/(165.5 − 101.6) = 21.8/63.9 = 0.3411, and T = 273.16 + 0.3411 × 327.34 = 273.16 + 111.7 = 384.8 K ≈ 111.7 °C.

Sanity check. 123.4 Ω lies between 101.6 Ω and 165.5 Ω, so T must lie between 273.16 K and 600.5 K. The value 384.8 K is about one-third of the way, matching the resistance ratio of 0.341.

Practical note. The linear law is only an approximation. Platinum resistance thermometers (Pt-100 has R = 100 Ω at 0 °C and α ≈ 3.85 × 10⁻³ K⁻¹) follow R = R₀(1 + At + Bt²) and are used as the interpolating standard of ITS-90 between 13.8 K and 961.78 °C.

Extra depth. Resistance thermometers exploit the fact that the resistivity of a metal rises with temperature because lattice vibrations scatter conduction electrons more strongly. Thermistors (semiconductors) have a negative temperature coefficient and are far more sensitive but strongly non-linear. A resistance thermometer is slower than a thermocouple but much more accurate and stable.

Conclusion: T ≈ 384.8 K (about 111.7 °C).

Concept. A good fixed point must be reproducible, sharply defined and independent of external conditions. The triple point of water occurs at a unique pressure and temperature, while the melting and boiling points depend on pressure and impurities.

Question. Answer the following: (a) Why is the triple point of water a standard fixed point and what is wrong with the ice point and steam point? (b) What is the other fixed point on the Kelvin scale? (c) Why is it 273.15 and not 273.16 in t_C = T − 273.15? (d) What is the triple point of water on an absolute scale whose unit interval equals that of Fahrenheit?

(a) Triple point versus ice and steam points

• The triple point of water (solid, liquid and vapour in equilibrium) occurs at exactly one temperature (273.16 K) and one pressure (611.73 Pa). It is therefore perfectly reproducible and requires no pressure control.
• The melting point of ice and boiling point of water vary markedly with pressure (the boiling point is about 100 °C at 1 atm but only about 70 °C on Mount Everest); dissolved impurities change both.
• The original Celsius scale used two fixed points, so the scale depended on the properties of the thermometric substance between them; the Kelvin scale needs only one fixed point plus absolute zero.

(b) The other fixed point

The other fixed point is absolute zero, 0 K, the lowest possible temperature, at which an ideal gas would exert zero pressure. The Kelvin scale is therefore defined by two points: 0 K and 273.16 K.

(c) Why 273.15

The Celsius scale was originally defined with the ice point at 0 °C. The ice point (pure water saturated with air at 1 atm) is slightly lower than the triple point, by 0.01 °C, because the triple point occurs at a low pressure (611.73 Pa) and without dissolved air. Hence 273.16 K corresponds to 0.01 °C, and 0 °C corresponds to 273.15 K. The relation t_C = T − 273.15 keeps the degree equal to 1 K and the ice point at 0 °C.

(d) Triple point on the Fahrenheit-size absolute scale

One Fahrenheit degree is 5/9 of a kelvin, so the number of degrees on this absolute scale (the Rankine scale) is 9/5 times the kelvin value:

T_R = (9/5)(273.16) = 491.688 ≈ 491.69

So the triple point of water is 491.69 degrees on the Rankine scale, and absolute zero is 0 °R = −459.67 °F.

Extra depth. The 1954 resolution of the General Conference on Weights and Measures adopted the triple point of water as the single fixed point defining the kelvin. In 2019 the kelvin was redefined through the Boltzmann constant, and the triple point of water became an experimentally determined value (273.1600 ± 0.0001 K). Water of known isotopic composition (VSMOW) is used for the triple-point cells.

Conclusion: (a) Triple point is unique and reproducible; (b) absolute zero 0 K; (c) ice point = 273.15 K is 0.01 K below the triple point; (d) 491.69.

Concept. In a constant-volume ideal-gas thermometer the pressure of a fixed mass of gas at constant volume is proportional to the absolute temperature: T = 273.16 K × (P/P_tr), where P_tr is the pressure at the triple point of water.

Question. Two ideal-gas thermometers A (oxygen) and B (hydrogen) give: at the triple point of water, P_A = 1.250 × 10⁵ Pa and P_B = 0.200 × 10⁵ Pa; at the normal melting point of sulphur, P_A = 1.797 × 10⁵ Pa and P_B = 0.287 × 10⁵ Pa. (a) Find the absolute temperature of the melting point of sulphur from A and B. (b) Why do the readings differ slightly and what procedure reduces the discrepancy?

(a) Calculation

T_A = 273.16 × (1.797/1.250) = 273.16 × 1.4376 = 392.7 K

T_B = 273.16 × (0.287/0.200) = 273.16 × 1.435 = 392.0 K

(b) Reason and remedy

The thermometers are not faulty. A real gas is not an ideal gas; the law P ∝ T holds exactly only in the limit of very low gas density, where intermolecular forces and finite molecular size are negligible. Oxygen and hydrogen deviate from ideal behaviour by different amounts at the same pressure (oxygen is more massive and more strongly interacting), so they give slightly different readings.

Procedure. Repeat the experiment with smaller and smaller amounts of gas, i.e. lower values of P_tr (say 0.8, 0.5, 0.2 of the original). Plot the measured T against P_tr for each gas and extrapolate the curve to P_tr → 0. All gases converge to the same intercept (the true ideal-gas temperature, about 392 K for sulphur), because at zero density every gas behaves ideally.

Percentage difference. (392.7 − 392.0)/392 × 100 = 0.18 %, so even the real gases agree to within 0.2 %, which is why the gas thermometer is the primary standard for low and moderate temperatures.

Extra depth. The constant-volume hydrogen thermometer was the standard until the 1950s; helium is now preferred because it is nearest to ideal behaviour down to very low temperatures (below 1 K the gas condenses, so helium-3 vapour-pressure and magnetic thermometers take over). Gas thermometers are bulky and slow, so they are used only to calibrate secondary thermometers such as platinum resistance thermometers.

Conclusion: (a) 392.7 K (A) and 392.0 K (B). (b) The gases are real, not ideal; use less gas (low pressure) and extrapolate to zero density.

Concept. A steel tape is graduated at 27.0 °C. At a higher temperature the tape itself expands, so each marked centimetre becomes longer than 1 cm and the tape under-reads. The true length is therefore larger than the reading by the factor (1 + αΔT).

Question. A steel tape 1 m long is correctly calibrated at 27.0 °C. The length of a steel rod measured by it on a hot day (45.0 °C) is 63.0 cm. What is the actual length of the rod on that day? What is its length at 27.0 °C? α_steel = 1.20 × 10⁻⁵ K⁻¹.

Given

• Calibration temperature T₁ = 27.0 °C; temperature of measurement T₂ = 45.0 °C; ΔT = 18.0 K
• Reading on the tape = 63.0 cm
• α = 1.20 × 10⁻⁵ K⁻¹

Step 1: actual length on the hot day

At 45 °C the divisions on the tape are stretched by the factor 1 + αΔT. The rod covers 63.0 tape divisions, so

L₄₅ = 63.0 × [1 + (1.20 × 10⁻⁵)(18.0)] = 63.0 × (1 + 2.16 × 10⁻⁴)

L₄₅ = 63.0 + 0.0136 = 63.0136 cm ≈ 63.01 cm

Step 2: length at 27.0 °C

The rod is steel too, so it has expanded by the same factor as the tape between 27 °C and 45 °C. The tape reading on the hot day is the rod length that it would have at the calibration temperature, because both the rod and the tape expanded together:

L₂₇ = L₄₅/(1 + αΔT) = 63.0136/1.000216 = 63.0 cm

Why the answers differ by so little

The fractional change is only αΔT = 2.16 × 10⁻⁴ = 0.0216 %, so the correction on 63 cm is 0.0136 cm = 0.136 mm, smaller than the least count of the tape (1 mm). It would matter only in precision engineering (e.g. a 100 m survey tape changes by about 2.2 cm for 18 K).

General rule. If the scale and the object are made of the same material, then the reading is the length at the temperature of calibration — a rod measured with a steel tape always reads its 27 °C length irrespective of the temperature. Only when the materials differ must the difference of the two expansion coefficients be used.

Extra depth. Surveyors use invar tapes (nickel-steel, α ≈ 1 × 10⁻⁶ K⁻¹) so that temperature has almost no effect; clock pendulums, metre standards and the pendulum of old precision clocks were also made of invar to avoid errors from expansion. In a railway track gaps are left between rails because 25 m of steel expands by 25 × 1.2 × 10⁻⁵ × 40 = 1.2 cm for a 40 K rise.

Conclusion: Actual length on the hot day = 63.0136 cm (≈ 63.01 cm); length at 27.0 °C = 63.0 cm.

Concept. A body contracts when cooled: ΔD = D α ΔT. To make the shaft slip into a hole that is 0.01 cm narrower than the shaft, the shaft diameter must shrink by 0.01 cm. The coefficient of linear expansion is the same for the shaft and wheel, so only the shaft’s contraction matters here.

Question. At 27 °C the outer diameter of a steel shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. The shaft is cooled with dry ice. At what temperature of the shaft does the wheel slip on the shaft? α_steel = 1.20 × 10⁻⁵ K⁻¹.

Given

• D_shaft(27 °C) = 8.70 cm; D_hole(27 °C) = 8.69 cm (wheel kept at 27 °C)
• α = 1.20 × 10⁻⁵ K⁻¹
• Required shaft temperature T so that D_shaft(T) = 8.69 cm

Solution

Required contraction of the shaft:

ΔD = 8.69 − 8.70 = −0.01 cm

Using ΔD = D α ΔT:

−0.01 = 8.70 × (1.20 × 10⁻⁵) × ΔT

ΔT = −0.01/(1.044 × 10⁻⁴) = −95.8 K

T = 27 − 95.8 = −68.8 °C ≈ −69 °C

Check against dry ice

Dry ice (solid CO₂) sublimes at −78.5 °C at 1 atm, so cooling the shaft with dry ice can easily reach −69 °C. The method is therefore practicable. A tiny additional margin (shaft cooled to about −78 °C) gives a comfortable clearance of 0.01 + 0.0010 cm.

Alternative approach. Heating the wheel until the hole diameter rises to 8.70 cm needs ΔT = 0.01/(8.69 × 1.2 × 10⁻⁵) = +95.9 K, i.e. 123 °C, and this is also commonly done (shrink fit). Cooling the shaft is chosen when heating the wheel would affect its heat treatment.

Why a shrink fit is strong. When the shaft returns to 27 °C it tries to expand again but is held by the wheel, creating a large radial pressure between the parts, so a firm joint results without any key or bolt.

Extra depth. The same principle is used to fit steel tyres on railway wheels (tyre heated, wheel cold), to fit bearing races, and to remove rivets. Because ΔD/D = αΔT, the required temperature change is independent of the actual diameter: only the fractional gap 0.01/8.70 = 1.15 × 10⁻³ matters.

Conclusion: The wheel slips on the shaft when the shaft is cooled to about −69 °C.

Concept. A hole in a sheet expands exactly as if it were made of the same metal as the sheet: every linear dimension, including the diameter of the hole, increases by the factor (1 + αΔT). The hole does not “shrink” as the metal expands into it.

Question. A hole is drilled in a copper sheet. Its diameter 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? α_copper = 1.70 × 10⁻⁵ K⁻¹.

Given

• D = 4.24 cm at 27.0 °C
• ΔT = 227 − 27 = 200 K
• α = 1.70 × 10⁻⁵ K⁻¹

Solution

ΔD = D α ΔT

ΔD = 4.24 × (1.70 × 10⁻⁵) × 200 = 1.4416 × 10⁻² cm

ΔD ≈ 1.44 × 10⁻² cm = 0.144 mm (increase)

The new diameter is 4.24 + 0.0144 = 4.2544 cm.

Why the hole expands

Imagine the hole filled with a disc of the same copper. On heating the disc would expand to a diameter D(1 + αΔT). Removing the disc leaves a hole of exactly this larger diameter. Atoms around the hole move apart just like atoms elsewhere, so the hole enlarges.

Area and volume

The area of the hole increases by ΔA/A = 2αΔT = 2 × 1.70 × 10⁻⁵ × 200 = 6.8 × 10⁻³ (0.68 %). A cavity expands in volume with γ = 3α, so a hollow copper vessel holds slightly more on heating.

Order of magnitude. The diameter changes by only 0.34 % of one percent: ΔD/D = αΔT = 3.4 × 10⁻³. Even for 200 K of heating, the change is about 0.14 mm, which is comparable to the clearance in a precision fit and hence matters in engineering design.

Extra depth. The rule “holes expand like the material” explains why a metal ring can be threaded over a ball at room temperature but not after the ball is heated (ball–ring experiment), and why a loose lid of a metal jar can be loosened by dipping the jar in hot water — the lid expands more than the glass jar.

More to know. Typical coefficients of linear expansion (×10⁻⁶ K⁻¹): aluminium 23, brass 19, copper 17, steel 12, glass 9, Pyrex 3.2, invar 0.9. A larger α means a larger change in every linear dimension for the same ΔT. A practical consequence is the bimetallic strip: two metals bonded together bend as the temperature changes, with the metal of larger α on the outer side of the curve when heated. Also remember that the percentage increase in the diameter equals the percentage increase in the circumference of the hole, so a ring fitted tightly on a rod may be loosened by heating the ring and tightened by cooling it.

Conclusion: The diameter of the hole increases by about 1.44 × 10⁻² cm (new diameter 4.2544 cm).

Concept. If the contraction of a rod or wire is prevented by rigid supports, a tension (thermal stress) develops. The wire tries to shorten by ΔL = LαΔT; the supports keep the length fixed, so the wire is stretched elastically by the same amount. Stress = Y × strain = Y α ΔT.

Question. 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 −39 °C, what tension is developed? Diameter = 2.0 mm, α = 2.0 × 10⁻⁵ K⁻¹, Y = 0.91 × 10¹¹ Pa.

Given

• L = 1.8 m; ΔT = 27 − (−39) = 66 K
• d = 2.0 mm so r = 1.0 × 10⁻³ m and A = πr² = 3.14 × 10⁻⁶ m²
• α = 2.0 × 10⁻⁵ K⁻¹; Y = 0.91 × 10¹¹ N m⁻²

Solution

Strain that the wire would have undergone freely: ΔL/L = αΔT = 2.0 × 10⁻⁵ × 66 = 1.32 × 10⁻³. Because the supports do not allow it, an equal tensile strain is produced:

Y = (F/A)/(ΔL/L) ⇒ F = Y A α ΔT

F = (0.91 × 10¹¹)(3.14 × 10⁻⁶)(2.0 × 10⁻⁵)(66)

F = 3.77 × 10² N ≈ 3.8 × 10² N

Check by stages

• Stress = Y α ΔT = 0.91 × 10¹¹ × 1.32 × 10⁻³ = 1.20 × 10⁸ Pa
• Force = stress × area = 1.20 × 10⁸ × 3.14 × 10⁻⁶ = 377 N

Is it safe? The stress 1.2 × 10⁸ Pa is well below the fracture strength of brass (about 4 × 10⁸ Pa) but of the order of its elastic limit, so the wire is strongly stretched. Notice that the length 1.8 m cancels out: thermal stress depends only on material (Y, α), ΔT and cross-section.

Physical meaning. The tension is about 38 kg-wt — like hanging a 38 kg mass from a 2 mm wire — and is generated purely by a 66 K fall in temperature. This explains why telephone wires are laid slack in summer and why railway tracks and bridges include expansion joints.

Extra depth. Thermal stress for a rod clamped at both ends: stress = YαΔT (compressive on heating, tensile on cooling). If one end is free no stress is produced, as in the next exercise (Ex 10.10). Large structures such as the Eiffel tower change in height by about 15 cm between winter and summer without stress because they are free to expand.

Conclusion: Tension developed ≈ 3.8 × 10² N.

Concept. When both ends of a composite rod are free, each part expands independently by its own ΔL = LαΔT and the total change is the sum. No stress arises because nothing prevents the expansion. A junction between unlike metals is stress-free only when the ends are unconstrained and the temperature is uniform.

Question. 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 at the junction? The ends are free. α_brass = 2.0 × 10⁻⁵ K⁻¹, α_steel = 1.2 × 10⁻⁵ K⁻¹.

Given

• L_brass = L_steel = 50 cm; ΔT = 250 − 40 = 210 K
• α_brass = 2.0 × 10⁻⁵ K⁻¹; α_steel = 1.2 × 10⁻⁵ K⁻¹
• Diameter 3.0 mm for both (does not affect ΔL)

Solution

ΔL_brass = 50 × (2.0 × 10⁻⁵) × 210 = 0.21 cm

ΔL_steel = 50 × (1.2 × 10⁻⁵) × 210 = 0.126 cm

ΔL_total = 0.21 + 0.126 = 0.336 cm ≈ 0.34 cm

Thermal stress at the junction?

No. The two rods expand by different amounts, but each is free to move; the junction simply moves along. The rods join end-to-end and each end of the combination is free, so there are no forces opposing the expansion. A thermal stress would arise only if either end were fixed (as in Ex 10.9) or if the rods were joined side by side as in a bimetallic strip, which bends instead.

Common mistake. The diameter (3.0 mm) is irrelevant for the change in length; it would be needed only for calculating a force.

Note on junction. The junction itself is a mathematical interface; the small step in radial dimension (brass expands radially 2.0 × 10⁻⁵ × 210 × 3.0 mm = 0.0126 mm while steel expands 0.0076 mm) causes a tiny mismatch in diameter but no longitudinal stress.

Extra depth. If instead the combined rod were clamped between rigid supports, both rods would be compressed by the same force F, and the junction would shift position. The equation to solve is F L/(A Y_brass) + F L/(A Y_steel) = (α_b + α_s) L ΔT, which gives a large force. Bimetallic strips (brass riveted to iron) bend when heated because the metal with the larger α lengthens more and sits on the outside of the bend; they are used in thermostats and fire alarms.

Conclusion: Total change in length ≈ 0.34 cm; no thermal stress develops at the junction.

Concept. Heating increases the volume of a liquid: V′ = V(1 + γΔT). The mass is unchanged, so the density falls by the same factor: ρ′ = ρ/(1 + γΔT). For small γΔT the fractional change in density is ≈ −γΔT.

Question. 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?

Given

• γ = 49 × 10⁻⁵ K⁻¹ = 4.9 × 10⁻⁴ K⁻¹
• ΔT = 30 K

Solution

Since density = mass/volume and the mass is constant:

ρ = m/V, ρ′ = m/V′ = ρ/(1 + γΔT)

Δρ/ρ = (ρ′ − ρ)/ρ = 1/(1 + γΔT) − 1 = −γΔT/(1 + γΔT)

Here γΔT = 4.9 × 10⁻⁴ × 30 = 1.47 × 10⁻², which is small compared with 1, so

Δρ/ρ ≈ −γΔT = −1.47 × 10⁻² ≈ −1.5 × 10⁻²

The negative sign shows a decrease in density of about 1.5 %. (The exact expression gives −1.449 × 10⁻², i.e. also −1.4 to −1.5 %.)

Comparison

Why density falls. Molecules vibrate with larger amplitudes when heated and push their neighbours slightly farther apart on average, so the same mass occupies more space. This is why hot liquid rises in a container and convection currents form.

Special case. Water is anomalous between 0 °C and 4 °C: it contracts on heating, reaching maximum density at 4 °C, and so lakes freeze from the top while the bottom stays at 4 °C. Glycerine has no such anomaly.

Extra depth. For solids the coefficients are related by γ = 3α (isotropic) and β = 2α for areas. For liquids only γ is meaningful because a liquid has no fixed shape; the apparent expansion measured in a glass vessel is γ_app = γ_liquid − γ_vessel.

More to know. The coefficient of volume expansion γ of common liquids (×10⁻⁴ K⁻¹): mercury 1.8, glycerine 4.9, alcohol 11, petrol 9.5, water about 2.1 near 20 °C. Because liquids expand much more than glass (γ_glass ≈ 0.27 × 10⁻⁴ K⁻¹), the apparent rise in a thermometer tube is given by γ_apparent = γ_liquid − γ_glass. The reduction in density with temperature drives natural convection in liquids and gases, which is the basis of sea breezes, ocean currents and the room-heater circulation.

Conclusion: Fractional change in density ≈ −1.5 × 10⁻² (a decrease of about 1.5 %).

Concept. The energy delivered by the machine is P t. If only a fraction η of it heats the block, the heat absorbed is Q = η P t = m c ΔT, which gives the temperature rise.

Question. 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 the 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.

Given

• P = 10 kW = 1.0 × 10⁴ W; t = 2.5 min = 150 s
• Fraction going into the block = 50 % = 0.5
• m = 8.0 kg = 8000 g; c = 0.91 J g⁻¹ K⁻¹

Solution

Heat absorbed by block, Q = 0.5 × P × t = 0.5 × 10⁴ × 150 = 7.5 × 10⁵ J

Q = m c ΔT ⇒ ΔT = Q/(m c) = 7.5 × 10⁵/(8000 × 0.91)

ΔT = 7.5 × 10⁵/7280 = 103 K

So the temperature of the block rises by about 103 °C (≈ 103 K).

Unit check

m c = 8000 g × 0.91 J g⁻¹ K⁻¹ = 7280 J K⁻¹ (or 8.0 kg × 910 J kg⁻¹ K⁻¹ = 7280 J K⁻¹). The answer ΔT = J/(J K⁻¹) = K, as it should be. Mixing grams with kg without adjusting the specific-heat unit is the most frequent source of error.

Interpretation

The power converted into heat in the block is 5 kW, i.e. 5000 J every second; this raises the temperature by 5000/7280 = 0.69 K s⁻¹. In 150 s the rise is 103 K. In practice a coolant would be needed, since an aluminium block starting at room temperature would then reach 130 °C, close to the softening range of many aluminium alloys.

Alternative form. Heat energy = work done against friction and cutting; the drill bit and chips remove part of the energy, which is why only 50 % is assumed here.

Extra depth. Specific heat capacity is the heat needed to raise the temperature of unit mass by 1 K: c = Q/(mΔT). Aluminium (0.91 J g⁻¹ K⁻¹) heats up faster than water (4.18 J g⁻¹ K⁻¹) for the same energy, which is why metal cookware warms quickly. The temperature rise is also reduced by heat conduction, so the real rise would be less than 103 K.

More to know. Power is energy per second, so 10 kW means 10 000 J of electrical energy every second. Efficiency here is 50 %, so the 5 kW is delivered to the block. If the block were also cooled by conduction to a vice, the actual rise would be smaller. The heat capacity of the block, C = mc = 7280 J K⁻¹, tells how much energy raises its temperature by 1 K; a block of copper (c = 0.39 J g⁻¹ K⁻¹) of the same mass would have heat capacity 3120 J K⁻¹ and would become hotter (240 K rise) under the same drill.

Conclusion: Rise in temperature of the block ≈ 103 °C.

Concept. The block loses heat as it cools from 500 °C to 0 °C; that heat melts ice at 0 °C. The maximum ice melts when the block cools all the way to 0 °C, and the heat lost equals m_ice × L_f.

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

Given

• m_Cu = 2.5 kg = 2500 g; ΔT = 500 − 0 = 500 K
• c_Cu = 0.39 J g⁻¹ K⁻¹
• L_f = 335 J g⁻¹

Solution

Heat lost by copper, Q = m c ΔT = 2500 × 0.39 × 500 = 4.875 × 10⁵ J

Heat gained by ice, Q = m_ice L_f ⇒ m_ice = 4.875 × 10⁵/335

m_ice = 1455 g ≈ 1.46 kg

A large ice block is at 0 °C, so the copper cannot cool below 0 °C and the ice cannot be heated above 0 °C; all the heat goes into melting. This gives the maximum mass of ice that melts (the final temperature is 0 °C).

Why “maximum”?

In practice some heat is lost to the surroundings and the ice block, being large, remains at 0 °C. Whatever heat flows from copper to the ice produces melting. We assume no loss to air, so the amount melted is the upper limit.

Cross-check

Heat released per kg of copper cooling by 500 K = 390 × 500 = 195 kJ, whereas melting 1 kg of ice needs 335 kJ. Therefore 2.5 kg of copper releases 487.5 kJ, enough for 487.5/335 = 1.455 kg of ice, which supports the answer.

Common error. Using the latent heat of vaporisation, or adding 4.18 J g⁻¹ K⁻¹ for warming the ice, is wrong; the ice is already at the melting point.

Extra depth. The specific latent heat of fusion of ice is large (335 J g⁻¹) compared with the sensible heat needed to warm water by 80 K (335 J g⁻¹); this is why a few ice cubes cool a drink so effectively. Calorimetry rests on the principle of conservation of energy: heat lost = heat gained.

Conclusion: Maximum ice melted ≈ 1.46 kg (about 1455 g).

Concept. In a calorimetric experiment, heat lost by the hot metal equals heat gained by the water and the calorimeter. The calorimeter is represented by its water equivalent (mass of water that would absorb the same heat).

Question. A 0.20 kg block of metal at 150 °C is dropped into a copper calorimeter (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 the answer greater or smaller than the true value?

Given

• Metal: m = 0.20 kg = 200 g; T_i = 150 °C; T_f = 40 °C
• Water: 150 cm³ = 150 g; calorimeter water equivalent = 25 g; both initially at 27 °C
• Specific heat of water = 4.18 J g⁻¹ K⁻¹

Solution

Heat lost by metal = 200 × c × (150 − 40) = 22000 c (J)

Heat gained = (150 + 25) × 4.18 × (40 − 27) = 175 × 4.18 × 13 = 9509.5 J

22000 c = 9509.5 ⇒ c = 0.432 J g⁻¹ K⁻¹ ≈ 0.43 J g⁻¹ K⁻¹ (= 430 J kg⁻¹ K⁻¹)

The metal could be iron-like (c ≈ 0.45) or brass (0.38).

Effect of heat loss

If some heat escapes to the surroundings, the water and calorimeter would gain less heat than the metal actually lost, and the final temperature would be lower than ideal. We assumed that all the heat lost by the metal went into the water; in reality the metal lost more. Our computed heat gain is therefore too small, so the computed specific heat is smaller than the actual value.

Check. Compare the two sides by ratio: the metal falls by 110 K while the water and calorimeter (175 g effective water) rise by 13 K, so c_metal/c_water = (175 × 13)/(200 × 110) = 0.103 and c_metal = 0.103 × 4.18 = 0.43 J g^-1 K^-1.

Reducing errors. Use an insulated (double-walled) calorimeter, stir gently, and transfer the hot block quickly.

Extra depth. The method of mixtures assumes an isolated system. Corrections are made by the cooling-correction (Regnault) method. The water equivalent of a calorimeter W = m_cal × c_cal/c_water; for copper (c = 0.39) a 64 g calorimeter has W = 25 g.

Conclusion: c ≈ 0.43 J g⁻¹ K⁻¹; with heat loss the computed value is smaller than the true value.

Concept. By the law of equipartition of energy, each quadratic degree of freedom contributes ½ k_B T per molecule, so ½ R T per mole. A monatomic gas has only 3 translational degrees of freedom; a diatomic molecule at room temperature also has 2 rotational degrees; at high temperature vibrational modes add 2 more (kinetic plus potential).

Question. Molar specific heats C_v (cal mol⁻¹ K⁻¹) at room temperature: H₂ 4.87, N₂ 4.97, O₂ 5.02, NO 4.99, CO 5.01, Cl₂ 6.17. These differ markedly from the monatomic value 2.92 cal mol⁻¹ K⁻¹. Explain the difference and infer something from the larger value for chlorine.

Why the difference

• Monatomic gas: only 3 translational degrees of freedom; C_v = (3/2)R = 3 × 0.5 × 1.987 ≈ 2.98 cal mol⁻¹ K⁻¹ (observed 2.92).
• The listed gases are diatomic. Besides translation, they can rotate about two axes perpendicular to the molecular axis (rotation about the bond axis has negligible moment of inertia). So f = 3 + 2 = 5 and C_v = (5/2)R = 4.97 cal mol⁻¹ K⁻¹.
• Observed values 4.87 – 5.02 match (5/2)R = 4.97, so at room temperature the vibrational modes are not excited.

Chlorine

The value for chlorine, 6.17 cal mol⁻¹ K⁻¹, is larger than 4.97 by about 1.2 cal mol⁻¹ K⁻¹, which is about 0.6 R. This shows that at room temperature a part of the vibrational degrees of freedom (the Cl–Cl bond is weak and has a low vibrational frequency, a large atomic mass) is already excited. If vibration were fully active, f would be 7 and C_v = (7/2)R = 6.95 cal mol⁻¹ K⁻¹. The observed 6.17 lies between 4.97 and 6.95, so vibration is partially excited.

Quantum explanation. A vibrational mode contributes only when kT is comparable to hν. For light molecules like H₂, N₂, O₂ the vibrational quantum is large (characteristic temperature 2200 – 6000 K), so these modes are “frozen” at 300 K. For Cl₂ the characteristic temperature is only ≈ 800 K, so the mode is partly active even at room temperature.

Extra depth. The temperature dependence of C_v of H₂ is a classic proof of quantum mechanics: C_v falls from (5/2)R to (3/2)R below about 50 K (rotation freezes) and rises above (5/2)R at around 1000 K when vibration is excited. Classical equipartition cannot explain this.

Conclusion: Diatomic molecules have extra rotational degrees of freedom, so C_v ≈ 5R/2 ≈ 4.97; chlorine’s larger value arises from partly active vibrational modes.

Concept. To lower the body temperature the child must lose heat Q = m c ΔT. Here the only mechanism is evaporation of sweat, each gram of which removes L = 580 cal. So the mass evaporated is Q/L, and dividing by the time gives the rate.

Question. A child running a temperature of 101 °F is given antipyrin, which increases evaporation of sweat. If the fever drops to 98 °F in 20 min, what is the average rate of extra evaporation? Mass of child = 30 kg; specific heat of human body ≈ that of water; latent heat of evaporation = 580 cal g⁻¹.

Given

• Fall in temperature = 101 − 98 = 3 °F = 3 × 5/9 = 1.667 °C (a difference: no 32)
• m = 30 kg = 30000 g; c = 1 cal g⁻¹ °C⁻¹
• L = 580 cal g⁻¹; time t = 20 min

Solution

Heat lost, Q = m c ΔT = 30000 × 1 × 1.667 = 5.0 × 10⁴ cal

Mass evaporated, m′ = Q/L = 5.0 × 10⁴/580 = 86.2 g

Rate = 86.2 g/20 min = 4.3 g min⁻¹

So the drug causes an extra evaporation of about 4.3 g of sweat per minute.

Why 3 °F is 5/3 °C

A temperature difference converts by the ratio of degree sizes, (5/9) °C per °F. Using (101 − 32) × 5/9 = 38.33 °C and (98 − 32) × 5/9 = 36.67 °C gives the same difference of 1.667 °C. Do not convert 3 °F as if it were a reading.

Magnitude check

A 30 kg child has a heat capacity of 30000 cal °C⁻¹ (30 kcal/°C), so cooling by 1.667 °C needs 50 kcal. Evaporation of 86 g of water at 580 cal g⁻¹ gives exactly this. Over 20 minutes the cooling power is 50000 cal/1200 s = 41.7 cal s⁻¹ ≈ 175 W, a realistic extra heat loss.

Biological significance. Sweat evaporates from the skin, taking latent heat from the body. Antipyretics reset the hypothalamic thermostat and dilate cutaneous vessels so the body sweats and loses heat faster.

Extra depth. Latent heat of vaporisation of water is very high (2260 J g⁻¹ at 100 °C, about 2430 J g⁻¹ = 580 cal g⁻¹ at skin temperature), so evaporation is an efficient cooling process. Humid weather slows evaporation and makes us feel hotter. Dogs pant, and humans sweat, for the same physical reason.

More to know. Normal human body temperature is 98.6 °F (37 °C). A fever of 101 °F is 38.3 °C. The body also loses heat by radiation, convection and conduction; in this idealised question only evaporation is considered. The 580 cal g⁻¹ value is the latent heat at skin temperature (about 35 °C); it is larger than 540 cal g⁻¹ at 100 °C because more energy is needed to evaporate at lower temperature. Heat stroke occurs when the evaporation cannot keep up with heat production.

Conclusion: Average extra evaporation ≈ 4.3 g min⁻¹ (≈ 86 g in 20 min).

Concept. Heat flows into the box by conduction through the thermocole walls at the rate H = K A (T₁ − T₂)/d. The ice melts and absorbs L_f per kg, so the mass melted in time t is m = H t/L_f.

Question. A cubical thermocole icebox of side 30 cm has wall thickness 5.0 cm. If 4.0 kg of ice is put in it, estimate the ice remaining after 6 h. Outside temperature = 45 °C, K_thermocole = 0.01 J s⁻¹ m⁻¹ K⁻¹; L_f = 335 × 10³ J kg⁻¹.

Given

• Side a = 30 cm = 0.30 m; total area of six faces A = 6a² = 6 × 0.09 = 0.54 m²
• d = 5.0 cm = 0.050 m; ΔT = 45 − 0 = 45 K (ice at 0 °C)
• K = 0.01 W m⁻¹ K⁻¹; t = 6 h = 21600 s

Solution

H = K A ΔT / d = (0.01 × 0.54 × 45)/0.050 = 4.86 W

Heat entering in 6 h: Q = H t = 4.86 × 21600 = 1.05 × 10⁵ J

Ice melted: m = Q/L_f = 1.05 × 10⁵/(3.35 × 10⁵) = 0.313 kg

Ice left = 4.0 − 0.313 = 3.69 kg ≈ 3.7 kg

Assumptions

• The outside area is used (0.54 m²); the inner area is slightly smaller (0.20 m side gives 0.24 m²), so the exact value is lower. The problem asks only for an estimate.
• Steady state; the inside of the wall is at 0 °C and the outside wall at 45 °C; heat flows only through the walls.
• The box contents are at 0 °C, so all heat entering goes into melting ice.

Estimate with the mean area. Taking the geometric mean of the inner (0.24 m²) and outer (0.54 m²) areas gives A ≈ 0.36 m². Then H = 3.24 W and melted ice = 0.21 kg, leaving about 3.8 kg. Both estimates show that most of the ice survives 6 h, which is why a thermocole box keeps food cold.

Why thermocole? It is expanded polystyrene with trapped air pockets; air is a very poor conductor (K ≈ 0.025), and the pockets are too small for convection. Hence K = 0.01 W m⁻¹ K⁻¹, about 40 000 times smaller than copper.

Extra depth. The thermal resistance of a slab is R = d/(KA); the rate of heat flow is H = ΔT/R, in complete analogy with Ohm’s law (I = V/R). Thermal resistances in series add up; those in parallel combine like conductances. Double walls with an air gap or vacuum (Thermos flask) lower K still further.

Conclusion: About 3.7 kg of ice remains after 6 h (≈ 0.31 kg melts).

Concept. In the steady state the heat conducted through the boiler base per second equals the heat absorbed by the boiling water per second, which equals (mass boiled per second) × L_v. The conduction formula then yields the temperature difference across the base.

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

Given

• A = 0.15 m²; d = 1.0 cm = 0.010 m
• Rate of boiling = 6.0 kg/min = 0.10 kg s⁻¹
• K = 109 W m⁻¹ K⁻¹; L_v = 2.256 × 10⁶ J kg⁻¹; water side at 100 °C

Solution

Heat required per second, H = (dm/dt) L_v = 0.10 × 2.256 × 10⁶ = 2.256 × 10⁵ W

H = K A (T₁ − T₂)/d ⇒ T₁ − T₂ = H d/(K A)

T₁ − T₂ = (2.256 × 10⁵ × 0.010)/(109 × 0.15) = 2256/16.35 = 138 K

T₁ = 100 + 138 = 238 °C

The flame in contact with the boiler base is at about 238 °C.

Interpretation

The base of the boiler is at the flame temperature on one face and at 100 °C (boiling water) on the other. The temperature drop of 138 K across a 1 cm brass plate is large because the heat flow is large (225 kW — compare a typical domestic gas ring of about 3 kW; the question is idealised). The estimate is the temperature of the flame just in contact with the boiler and not of the hottest part of the flame (which exceeds 1000 °C).

Power check. 2.256 × 10⁵ W is 225.6 kW, much more than a domestic burner; the numbers in the problem were chosen for easy calculation. Do not worry about realism in the exam, but know the method.

Extra depth. Thermal resistance of the base R = d/(KA) = 0.01/(109 × 0.15) = 6.1 × 10⁻⁴ K W⁻¹, so the drop ΔT = H R = 2.256 × 10⁵ × 6.1 × 10⁻⁴ = 138 K. In real boilers a layer of scale on the water side (poor conductor) can raise the metal temperature dangerously.

Conclusion: The flame in contact with the boiler is at about 238 °C.

Concept. Radiation, conduction and convection all follow from basic laws: Kirchhoff’s law (good absorbers are good emitters), the dependence of heat conduction on thermal conductivity, and Stefan’s law for black-body radiation.

Question. 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 calibrated for ideal black-body radiation gives too low a value for a red-hot iron piece in the open but the correct value in a furnace; (d) the earth without its atmosphere would be inhospitably cold; (e) steam-based heating systems warm a building more efficiently than hot-water systems.

(a) Large reflectivity means poor emitter

For a surface, absorptivity a + reflectivity r + transmittance t = 1; for an opaque body t = 0 so a = 1 − r. A highly reflecting body has small a. By Kirchhoff’s law the emissive power of a body at a given temperature is proportional to its absorptivity, e = a E_b. Therefore small a means small emissivity: it is a poor emitter (as well as a poor absorber). This is why a silvered Thermos flask wall minimises radiation loss.

(b) Brass tumbler feels colder

Brass has high thermal conductivity (≈ 109 W m⁻¹ K⁻¹) compared with wood (≈ 0.1 – 0.2). When we touch it, heat flows quickly from our hand to the brass and is conducted away, so the skin temperature falls and the sensation is of cold. Wood conducts so poorly that the surface warms to the skin temperature, and little heat is drawn from the hand. Both are at the same room temperature.

(c) Optical pyrometer

An optical pyrometer is calibrated for a black body, which emits the maximum possible radiation (e = 1) at a given temperature. A red-hot iron piece in the open has emissivity less than 1; the radiation it emits is smaller than that of a black body at the same temperature, so the pyrometer matches it to a lower temperature and gives too low a reading. In a furnace the iron piece is in equilibrium with the cavity (a near-perfect black body), and the radiation from the piece (including that reflected from the walls) is the black-body radiation; hence the reading is correct.

(d) Earth without the atmosphere

Earth’s surface absorbs solar radiation (short wavelengths) and re-emits it as infrared radiation. The atmosphere (CO₂, water vapour, methane, ozone) absorbs this infrared radiation and re-radiates part of it back to the surface (the greenhouse effect). Without the atmosphere, all the infrared would escape to space; the average temperature would be about −18 °C instead of +15 °C, so the Earth would be inhospitably cold.

(e) Steam heating versus hot water

Steam at 100 °C has a large latent heat of vaporisation (2256 J g⁻¹) in addition to its sensible heat. When steam condenses in radiators it releases 2256 J g⁻¹, while the same mass of hot water cooling by, say, 30 K gives only 4.18 × 30 = 125 J g⁻¹. Hence a given mass of steam supplies about 18 times as much heat as hot water; the steam system is more efficient.

Extra depth. Key laws: Kirchhoff’s law e/a = E_b(λ, T); Stefan–Boltzmann law E = σ T⁴ with σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴; Wien’s displacement law λ_m T = 2.9 × 10⁻³ m K. Pyrometers use these laws to measure temperatures above 1000 °C without contact.

Conclusion: (a) e ∝ a = 1 − r; (b) brass conducts heat away from the hand; (c) e < 1 outside, black body in a furnace; (d) no greenhouse effect; (e) steam carries latent heat.

Concept. Newton’s law of cooling: the rate of fall of temperature is proportional to the excess of temperature over the surroundings: −dT/dt = k (T − T₀). For small differences, the average temperature excess is used.

Question. 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 surroundings are at 20 °C.

Given

• Surroundings T₀ = 20 °C
• Case 1: 80 °C → 50 °C in 5 min; mean temperature = 65 °C, excess = 45 °C
• Case 2: 60 °C → 30 °C in t min; mean temperature = 45 °C, excess = 25 °C

Solution (average-temperature form)

(T₁ − T₂)/t = k [(T₁ + T₂)/2 − T₀]

(80 − 50)/5 = k (65 − 20) ⇒ 6 = 45 k ⇒ k = 0.1333 min⁻¹

(60 − 30)/t = k (45 − 20) = 0.1333 × 25 = 3.33

t = 30/3.33 = 9 min

Exact (exponential) solution

Integrating dT/dt = −k (T − T₀) gives ln[(T₁ − T₀)/(T₂ − T₀)] = k t. From case 1: k = (1/5) ln(60/30) = (1/5)(0.693) = 0.1386 min⁻¹. For case 2: t = (1/k) ln(40/10) = 1.386/0.1386 = 10 min.

Which one to give? The NCERT answer is 9 minutes (average-temperature approximation). In the exact exponential solution the time is exactly 10 minutes, because cooling from 60 to 30 °C takes the excess from 40 to 10 (a factor 4 = 2²), which needs two “half-lives” of 5 minutes. Quote 9 min as the textbook answer and mention 10 min as the more accurate value.

Sanity check. The body cools more slowly in the second case because the excess over surroundings is smaller; in 5 min it falls by only about 16 °C. A 30 °C fall therefore takes longer than 5 min, so 9 – 10 min is reasonable.

Extra depth. Newton’s law holds approximately for small excess temperatures (about 30 K) with forced convection; for natural convection and radiation the exponent is higher (Stefan’s law gives a T⁴ dependence). It underlies the determination of the time of death in forensic science and the cooling curves in calorimetry.

Conclusion: Time ≈ 9 min (exact exponential solution: 10 min).