Exercises
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.
Answer
Conversion formulas: TC = TK − 273.15, TF = (9/5)TC + 32
For Neon (24.57 K)
- TC = 24.57 − 273.15 = −248.58 °C
- TF = (9/5)(−248.58) + 32 = −447.44 + 32 = −415.44 °F
For Carbon Dioxide (216.55 K)
- TC = 216.55 − 273.15 = −56.60 °C
- TF = (9/5)(−56.60) + 32 = −101.88 + 32 = −69.88 °F
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 TA and TB?
Answer
TA/200 = TK/273.16 and TB/350 = TK/273.16
Dividing: 7TA = 4TB
3: Resistance thermometer: R = R₀[1 + α(T − T₀)]. R = 101.6 Ω at 273.16 K; R = 165.5 Ω at 600.5 K. Find T when R = 123.4 Ω.
Answer
α = (165.5/101.6 − 1) / (600.5 − 273.16) = 0.629/327.34 = 1.92 × 10⁻³ K⁻¹
123.4 = 101.6[1 + 1.92 × 10⁻³ × (T − 273.16)]
T − 273.16 = 0.2146 / (1.92 × 10⁻³) = 111.8 K
T ≈ 384.9 K
4: Questions about temperature scales.
Answer
(a) Triple point is unique (single T and P), independent of external conditions. Melting/boiling points depend on pressure — unreliable as standards.
(b) Absolute zero (0 K) is the other fixed point.
(c) The melting point of ice (0 °C) = 273.15 K (not 273.16 K), so tc = T − 273.15.
(d) Triple point on Fahrenheit-sized Kelvin scale = 273.16 × 9/5 = 491.69
5: Two ideal gas thermometers A (O₂) and B (H₂). Find absolute temperature of melting point of sulphur.
Answer
T = 273.16 × (P/Ptriple)
(a) Thermometer A: T = 273.16 × (1.797/1.250) = 392.69 K
Thermometer B: T = 273.16 × (0.287/0.200) = 391.98 K
(b) Slight difference because gases are not perfectly ideal. Extrapolate to zero pressure for accurate result.
6: Steel tape 1 m calibrated at 27 °C; measures rod as 63.0 cm at 45 °C. Actual length at 45 °C and 27 °C? α = 1.20 × 10⁻⁵ K⁻¹.
Answer
Tape length at 45 °C = 100 × [1 + 1.20 × 10⁻⁵ × 18] = 100.0216 cm per 100 cm
Actual length at 45 °C = 63.0 × (100.0216/100) = 63.014 cm
Length at 27 °C = 63.014 / [1 + 1.20 × 10⁻⁵ × 18] = 63.0 cm
7: Steel wheel (hole 8.69 cm) to fit on shaft (8.70 cm) at 27 °C. Cool shaft until wheel slips. α = 1.20 × 10⁻⁵ K⁻¹.
Answer
ΔT = (8.69 − 8.70)/(8.70 × 1.20 × 10⁻⁵) = −0.01/1.044 × 10⁻⁴ = −95.8 K
T = 300 − 95.8 = 204.2 K = −68.8 °C ≈ −69 °C
8: Copper sheet hole diameter 4.24 cm at 27 °C, heated to 227 °C. Change in hole diameter? αCu = 1.70 × 10⁻⁵ K⁻¹.
Answer
Δd = d₀ · α · ΔT = 4.24 × 1.70 × 10⁻⁵ × 200 = 1.44 × 10⁻² cm
The hole diameter increases (hole expands like a disc of same material).
9: Brass wire 1.8 m, diameter 2.0 mm, cooled from 27 °C to −39 °C. Find tension. α = 2.0 × 10⁻⁵ K⁻¹, Y = 0.91 × 10¹¹ Pa.
Answer
A = π × (1.0 × 10⁻³)² = 3.14 × 10⁻⁶ m²
Strain = α·ΔT = 2.0 × 10⁻⁵ × 66 = 1.32 × 10⁻³
Tension = Y × strain × A = 0.91 × 10¹¹ × 1.32 × 10⁻³ × 3.14 × 10⁻⁶ = 3.77 × 10² N
10: Brass (50 cm) + steel (50 cm) joined rod, heated from 40 °C to 250 °C. Total change in length? Thermal stress? αbrass = 2.0 × 10⁻⁵, αsteel = 1.2 × 10⁻⁵ K⁻¹.
Answer
ΔLbrass = 0.50 × 2.0 × 10⁻⁵ × 210 = 2.1 × 10⁻³ m
ΔLsteel = 0.50 × 1.2 × 10⁻⁵ × 210 = 1.26 × 10⁻³ m
Total ΔL = 3.36 × 10⁻³ m = 0.336 cm
No thermal stress — ends are free to expand.
11: Glycerine γ = 49 × 10⁻⁵ K⁻¹. Fractional change in density for 30 °C rise?
Answer
Fractional change in density ≈ −γΔT = −49 × 10⁻⁵ × 30 = −1.47 × 10⁻²
Density decreases by 1.47%.
12: 10 kW drill on 8.0 kg aluminium block for 2.5 min. 50% energy heats block. ΔT? cAl = 0.91 J g⁻¹ K⁻¹.
Answer
Q = 0.5 × 10,000 × 150 = 7.5 × 10⁵ J
ΔT = Q/(mc) = 7.5 × 10⁵ / (8000 × 0.91) = 103 °C
13: 2.5 kg copper block at 500 °C placed on ice. Maximum ice melted? cCu = 0.39 J g⁻¹ K⁻¹, L = 335 J g⁻¹.
Answer
Q = 2500 × 0.39 × 500 = 4.875 × 10⁵ J
mice = Q/L = 4.875 × 10⁵ / 335 = 1455 g ≈ 1.5 kg
14: Metal block 0.20 kg at 150 °C in calorimeter (water equiv. 0.025 kg) + 150 g water at 27 °C. Final T = 40 °C. Find specific heat.
Answer
200 × c × 110 = 175 × 1 × 13
c = 2275/22000 = 0.103 J g⁻¹ K⁻¹
If heat losses occur, actual c is greater than calculated.
15: Explain why diatomic gas molar specific heats (~5 cal/mol K) exceed monatomic (~2.92 cal/mol K). Why is Cl₂ larger?
Answer
Monatomic: 3 translational DoF
⇒ CV = 3R/2 ≈ 2.98 cal/mol K
Diatomic (H₂, N₂, O₂): 3 translational + 2 rotational DoF
⇒ CV = 5R/2 ≈ 4.97 cal/mol K
Cl₂ is larger (~6.17) because its vibrational modes are also active at room temperature (Cl–Cl bond has low frequency), adding 2 more DoF
⇒ CV ≈ 7R/2 ≈ 6.95 cal/mol K.
16: Child's fever reduced 101°F → 98°F in 20 min by evaporation. Mass = 30 kg, L = 580 cal/g. Rate of evaporation?
Answer
ΔT = 3°F = 5/3 °C; Q = 30,000 × 1 × 5/3 = 50,000 cal
Rate of evaporation = (50,000/20) / 580 = 4.3 g/min
17: Thermacole icebox 30 cm side, 5.0 cm thick. Outside 45 °C, inside 0 °C. Ice remaining after 6 h? k = 0.01 J s⁻¹ m⁻¹ K⁻¹.
Answer
A = 6 × (0.30)² = 0.54 m²
Q/t = kAΔT/d = 0.01 × 0.54 × 45 / 0.05 = 4.86 J/s
Total Q = 4.86 × 21,600 = 1.05 × 10⁵ J
Ice melted = 1.05 × 10⁵ / (335 × 10³) = 0.313 kg
Ice remaining = 4.0 − 0.313 ≈ 3.687 kg
18: Brass boiler: area 0.15 m², thickness 1.0 cm, boils 6.0 kg water/min. Flame temperature? kbrass = 109 J s⁻¹ m⁻¹ K⁻¹.
Answer
Q/t = 6.0 × 2256 × 10³ / 60 = 2.256 × 10⁵ J/s
ΔT = (Q/t) × d / (kA) = 2.256 × 10⁵ × 0.01 / (109 × 0.15) = 138 °C
Flame temperature = 100 + 138 = 238 °C
19: Explain (a) large reflectivity → poor emitter; (b) brass colder than wood; (c) pyrometer error in open; (d) cold Earth without atmosphere; (e) steam heating more efficient.
Answer
(a) Kirchhoff's law: absorptivity = emissivity. High reflectivity
⇒ low absorptivity
⇒ low emissivity
⇒ poor emitter.
(b) Brass is a good thermal conductor — rapidly draws heat from skin
⇒ feels colder. Wood is a poor conductor
⇒ feels warmer.
(c) Pyrometer calibrated for ε = 1 (black body). Iron in open has ε < 1
⇒ reads low. Inside furnace, iron and enclosure form effective black body
⇒ correct reading.
(d) Without the greenhouse effect of the atmosphere, Earth's surface would radiate all heat to space, giving average temperature ≈ −18 °C.
(e) Steam at 100 °C releases latent heat of condensation (2260 J/g) in addition to sensible heat, providing far more energy per unit mass than hot water.
20: Body cools 80 °C → 50 °C in 5 min. Time to cool 60 °C → 30 °C? Surroundings = 20 °C.
Answer
Newton's cooling: rate ∝ (Tavg − T₀)
Case 1: avg = 65 °C, excess = 45 °C, rate = 6 °C/min
⇒ k = 6/45 = 2/15 min⁻¹
Case 2: avg = 45 °C, excess = 25 °C, rate = (2/15) × 25 = 10/3 °C/min
Time = 30 ÷ (10/3) = 9 minutes
Important Extra Questions
Extra Q1: What is the difference between heat and temperature? Why is touch unreliable for measuring temperature?
Answer:
Temperature measures average kinetic energy of molecules (intensive property). Heat is energy in transit due to temperature difference (extensive property).
Touch is unreliable because it senses the rate of heat flow to/from the skin, which depends on thermal conductivity, not just temperature. Brass and wood at 20 °C feel very different because brass conducts heat away much faster.
Extra Q2: Steel rail 30 m long at 20 °C. What gap is needed for maximum temperature of 60 °C? α = 1.2 × 10⁻⁵ K⁻¹.
Answer:
ΔL = L·α·ΔT = 30 × 1.2 × 10⁻⁵ × 40 = 1.44 × 10⁻² m = 1.44 cm
A gap of at least 1.44 cm must be left at each joint.
Extra Q3: Explain latent heat. Why does ice-water mixture stay at 0 °C when heat is supplied?
Answer:
Latent heat is heat absorbed/released during phase change without temperature change. During melting, heat breaks intermolecular bonds (ice crystal structure), not increasing kinetic energy. Temperature stays at 0 °C until all ice melts. Lfusion(water) = 334 J/g; Lvaporization = 2260 J/g.
Extra Q4: State Newton's Law of Cooling. Under what conditions does it apply?
Answer:
dT/dt = −k(T − T₀)
Rate of cooling is proportional to the excess temperature above surroundings.
Conditions: Small temperature excess; primarily convective/radiative cooling; surroundings at constant temperature.
Solution: T(t) = T₀ + (Ti − T₀)e−kt
Extra Q5: Explain the anomalous expansion of water and its ecological significance.
Answer:
Water contracts normally from 100 °C to 4 °C (maximum density at 4 °C), then expands from 4 °C to 0 °C — anomalous expansion due to formation of open hydrogen-bonded ice-like clusters.
Ecological significance: In winter, surface water cools to 4 °C and sinks; further-cooled water (0–4 °C) is lighter and stays at top, freezing first. Ice insulates below, maintaining liquid water at ~4 °C where aquatic life survives.