Heat and temperature
Temperature tells how hot or cold a body is. It decides which way heat flows: always from higher to lower temperature. Heat is the energy that flows from one body to another only because of a temperature difference. Unit of heat: joule (J); older unit: calorie (1 cal = 4.186 J).
Temperature scales: Celsius (ice 0, steam 100), Fahrenheit (ice 32, steam 212) and Kelvin (absolute scale). Relations: tF = (9/5) tC + 32 and T = tC + 273.15. Absolute zero (0 K = −273.15 °C) is the lowest possible temperature.
An ideal gas follows PV = μRT. A constant-volume gas thermometer uses the fact that pressure ∝ T, which is how the absolute scale is defined.
Thermal expansion of solids, liquids and gases
When heated, particles vibrate more strongly and push each other slightly apart, so most bodies grow.
- Linear: ΔL/L = α ΔT
- Area: ΔA/A = β ΔT
- Volume: ΔV/V = γ ΔT (γ also called αV)
For a solid, β = 2α and γ = 3α. Proof idea for γ: take a cube of side l. New side = l(1 + αΔT). New volume = l³(1 + αΔT)³ ≈ l³(1 + 3αΔT) since αΔT is tiny. So ΔV/V = 3αΔT.
Metals expand more than glass. Copper: α ≈ 1.7 × 10⁻⁵ K⁻¹; glass ≈ 0.9 × 10⁻⁵ K⁻¹. Liquids expand more than solids, and gases the most. For an ideal gas at constant pressure, γ = 1/T (at 0 °C, γ = 1/273 K⁻¹ = 3.7 × 10⁻³ K⁻¹).
Thermal stress: if a rod is fixed at both ends and heated, it cannot expand, so a stress Y α ΔT builds up. This is why rails, bridges and pipelines have expansion gaps or loops.
Odd (anomalous) expansion of water
Water behaves differently between 0 °C and 4 °C: it contracts on heating. Above 4 °C it expands normally. So water has its maximum density at 4 °C.
In winter, the top of a lake cools. Water at 4 °C, being densest, sinks to the bottom. Colder water (below 4 °C) stays on top and freezes into ice, which floats. Ice is a poor conductor of heat, so the water below stays near 4 °C and fish survive.
Water also expands about 9% when it freezes, which is why water pipes can burst in very cold places.
Specific heat capacity
Heat capacity S = ΔQ/ΔT: heat needed to raise the whole body by 1 K.
Specific heat capacity c = ΔQ/(m ΔT): heat needed to raise 1 kg by 1 K. Unit: J kg⁻¹ K⁻¹. So Q = m c ΔT.
Molar specific heat C = ΔQ/(μ ΔT) for μ moles. Unit: J mol⁻¹ K⁻¹.
Water has a very high specific heat (4186 J/kg K). Iron: about 450; aluminium: about 900. This is why water is used in car radiators and hot-water bags, and why sea breezes happen.
Cp and Cv for gases
For solids and liquids the volume barely changes, so there is one specific heat. For a gas it matters how you heat it:
- Cv: molar specific heat at constant volume. All heat goes to raise internal energy.
- Cp: molar specific heat at constant pressure. The gas also expands and does work, so it needs extra heat.
So Cp > Cv, and for an ideal gas Cp − Cv = R (Mayer's relation, R = 8.314 J mol⁻¹ K⁻¹). Their ratio γ = Cp/Cv is 5/3 for monatomic gases and 7/5 for diatomic gases like air.
Calorimetry
Calorimetry means measuring heat. A calorimeter is a copper or aluminium cup with a stirrer, kept in a wooden jacket with insulation so almost no heat escapes.
Principle of calorimetry: in an isolated system, heat lost by hot bodies = heat gained by cold bodies. It is just conservation of energy.
Method of mixtures to find c of a metal: heat a metal piece of mass m₁ to T₁, drop it into water (mass m₂) in a calorimeter (mass m₃, specific heat c₃) at T₂. Measure the final temperature T. Then m₁c₁(T₁ − T) = (m₂cw + m₃c₃)(T − T₂). Solve for c₁.
Change of state and latent heat
Solid ⇄ liquid ⇄ gas are the changes of state. Melting (fusion) happens at the melting point; boiling at the boiling point. At these points, heat is absorbed but the temperature does not rise until the change is complete. Both states exist together.
Latent heat L = heat per unit mass needed to change the state without changing the temperature: Q = m L. Unit: J/kg.
- Latent heat of fusion of ice: Lf = 3.34 × 10⁵ J/kg
- Latent heat of vaporisation of water: Lv = 22.6 × 10⁵ J/kg
The heating curve (temperature vs heat) of ice has sloping parts (warming) and flat parts (melting at 0 °C and boiling at 100 °C). The flat boiling part is very long because Lv is large.
The boiling point rises with pressure (pressure cooker cooks faster) and falls at low pressure (water boils below 100 °C on mountains). The melting point of ice falls a little with pressure: a wire with weights passes slowly through an ice block and the ice refreezes above it (regelation). Sublimation is solid to gas directly (camphor, dry ice). The triple point of water (273.16 K, 0.006 atm) is where all three states exist together.
Try it: the melting ice watch
Put crushed ice in a steel glass, stir, and read a kitchen thermometer every minute. Write the readings. The number stays near 0 °C until the last ice melts, then it climbs. That flat part is latent heat of fusion at work. In the 3D, step 6, drag Q to 42 kJ and then to 376 kJ: the temperature stays at 0 °C the whole way.
Key formulas and definitions
- t_F = (9/5) t_C + 32; T = t_C + 273.15
- ΔL = α L ΔT; ΔA = β A ΔT; ΔV = γ V ΔT
- β = 2α; γ = 3α; ideal gas γ = 1/T
- Thermal stress = Y α ΔT
- Q = m c ΔT; heat capacity S = m c
- Cp − Cv = R (per mole)
- Heat lost = heat gained
- Q = m L
Worked examples
1. Convert 37 °C to °F and to K.
Step 1: t_F = (9/5) × 37 + 32 = 66.6 + 32 = 98.6 °F. Step 2: T = 37 + 273.15 = 310.15 K. Answer: 98.6 °F, 310.15 K.
2. A copper rod is 1 m long at 20 °C. How much longer is it at 320 °C? (α = 1.7 × 10⁻⁵ K⁻¹)
Step 1: ΔT = 300 K. Step 2: ΔL = αLΔT = 1.7 × 10⁻⁵ × 1 × 300. Answer: 5.1 × 10⁻³ m = 5.1 mm.
3. A brass sheet (α = 1.9 × 10⁻⁵ K⁻¹) has area 0.5 m². Find the increase in area for a 100 K rise.
Step 1: β = 2α = 3.8 × 10⁻⁵ K⁻¹. Step 2: ΔA = βAΔT = 3.8 × 10⁻⁵ × 0.5 × 100. Answer: 1.9 × 10⁻³ m² = 19 cm².
4. How much heat raises 2 kg of water from 25 °C to 75 °C? (c = 4186 J/kg K)
Step 1: ΔT = 50 K. Step 2: Q = mcΔT = 2 × 4186 × 50. Answer: 4.186 × 10⁵ J ≈ 419 kJ.
5. A 0.2 kg aluminium ball at 100 °C is dropped into 0.3 kg of water at 20 °C. Find the final temperature. (c_Al = 900, c_w = 4186 J/kg K, ignore the calorimeter)
Step 1: Heat lost by Al = 0.2 × 900 × (100 − T) = 180(100 − T). Step 2: Heat gained by water = 0.3 × 4186 × (T − 20) = 1255.8(T − 20). Step 3: Set equal: 18000 − 180T = 1255.8T − 25116. Step 4: 43116 = 1435.8T. Answer: T ≈ 30 °C.
6. A steel rail fixed at both ends is heated by 40 K. Find the thermal stress. (Y = 2 × 10¹¹ Pa, α = 1.2 × 10⁻⁵ K⁻¹)
Step 1: Stress = YαΔT. Step 2: = 2 × 10¹¹ × 1.2 × 10⁻⁵ × 40. Answer: 9.6 × 10⁷ Pa. That is huge, so gaps are left between rails.
7. How much heat turns 0.5 kg of ice at −10 °C into water at 20 °C? (c_ice = 2100, c_w = 4186 J/kg K, L_f = 3.34 × 10⁵ J/kg)
Step 1: Warm ice −10 → 0 °C: 0.5 × 2100 × 10 = 10500 J. Step 2: Melt: 0.5 × 3.34 × 10⁵ = 167000 J. Step 3: Warm water 0 → 20 °C: 0.5 × 4186 × 20 = 41860 J. Step 4: Add: 10500 + 167000 + 41860. Answer: 219360 J ≈ 2.19 × 10⁵ J.
8. 10 g of steam at 100 °C passes into 1 kg of water at 20 °C. Find the final temperature. (L_v = 2.26 × 10⁶ J/kg, c = 4186 J/kg K)
Step 1: Heat given by steam = condensing + cooling = 0.01 × 2.26 × 10⁶ + 0.01 × 4186 × (100 − T) = 22600 + 41.86(100 − T). Step 2: Heat taken by water = 1 × 4186 × (T − 20). Step 3: 22600 + 4186 − 41.86T = 4186T − 83720. Step 4: 110506 = 4227.86T. Answer: T ≈ 26.1 °C.
Common mistakes
- Thinking heat and temperature are the same. A bucket of warm water has more heat than a cup of hot tea, though its temperature is lower.
- Using γ = α or β = α. For a solid, β = 2α and γ = 3α.
- Forgetting the latent heat step in mixing problems with ice or steam. The flat parts of the curve need Q = mL.
- Thinking water always expands on heating. Between 0 and 4 °C it contracts.