Temperature, heat and temperature scales
A cup of tea feels hotter than a glass of water at room temperature, and heat flows from the tea to the cup. Link to earlier classes : heat and temperature of class 7 and 9 are now defined more carefully.
Temperature is a measure of how hot or cold a body is. Heat is the energy that flows between two bodies because of a temperature difference; it flows from higher to lower temperature. Unit of heat : joule (J).
Rule 1: Temperature scales — F = (9 ÷ 5) C + 32 and K = C + 273.15 (C = degree Celsius, F = degree Fahrenheit, K = kelvin, the SI unit). The triple point of water is 273.16 K.
- Easy : 37 °C = (9 ÷ 5) × 37 + 32 = 98.6 °F.
- Medium : 50 °C = 122 °F = 323 K.
- Exam : At −40 the Celsius and Fahrenheit readings are equal : C = (9 ÷ 5) C + 32 ⇒ C = −40.
Heat is energy in transit; temperature is a state; K = C + 273.
Calling temperature "the amount of heat in a body".
Memory clue : "Heat Hops, Temperature Tells."
Spot the difference
| Quantity | Meaning | Unit |
|---|---|---|
| Heat | energy that flows | joule |
| Temperature | degree of hotness | kelvin |
The ideal-gas equation
A football pumped in the cold morning feels softer than in the afternoon sun. Link to earlier classes : Boyle's and Charles' laws of class 9 are now joined into one equation.
An ideal gas obeys one relation between pressure, volume and absolute temperature at all conditions : it is a good model for real gases at low pressure and high temperature.
Rule 1: Ideal-gas equation — P V = n R T (P = pressure, V = volume, n = number of moles, R = 8.314 J mol⁻¹ K⁻¹, T = temperature in kelvin). For a fixed amount of gas, P₁V₁ ÷ T₁ = P₂V₂ ÷ T₂.
- Easy : 1 mol at 273.15 K and 1 atm occupies V = nRT ÷ P = 8.314 × 273.15 ÷ 101 325 = 0.0224 m³ = 22.4 litre.
- Medium : 2 mol at 300 K in a 10 litre vessel gives P = 2 × 8.314 × 300 ÷ 0.010 = 4.99 × 10⁵ Pa.
- Exam : A gas at 27 °C and 1 atm is compressed to half its volume while the temperature rises to 127 °C. P₂ = P₁ × (V₁ ÷ V₂) × (T₂ ÷ T₁) = 1 × 2 × 400 ÷ 300 = 2.67 atm.
Always use kelvin in the gas laws.
Putting Celsius temperature into P V = n R T.
Memory clue : "Please Visit New Restaurant Today : P V = n R T."
Spot the difference
| Law | Constant | Relation |
|---|---|---|
| Boyle | T | P ∝ 1 ÷ V |
| Charles | P | V ∝ T |
| Gay-Lussac | V | P ∝ T |
Thermal expansion
Gaps are left between railway tracks, and a tight metal lid opens when held under hot water. Link to earlier classes : expansion on heating of class 7 is now measured.
Most solids, liquids and gases expand when heated. The coefficient of linear expansion α is the fractional change in length per kelvin.
Rule 1: Linear, area and volume expansion — ΔL = α L ΔT. The area coefficient is about 2α and the volume coefficient is γ = 3α.
- Easy : A steel rod (α = 1.2 × 10⁻⁵ K⁻¹) of length 10 m heated by 30 K expands by 1.2 × 10⁻⁵ × 10 × 30 = 3.6 × 10⁻³ m = 3.6 mm.
- Medium : Copper has α = 1.7 × 10⁻⁵ K⁻¹, so γ = 5.1 × 10⁻⁵ K⁻¹.
- Exam : A steel rod fixed between two rigid walls is heated by 50 K. The thermal stress = Y α ΔT = 2 × 10¹¹ × 1.2 × 10⁻⁵ × 50 = 1.2 × 10⁸ Pa.
Rule 2: Anomalous expansion of water — Water contracts as it is heated from 0 °C to 4 °C, so its density is the highest at 4 °C. That is why a lake freezes from the top and fish survive underneath.
- Easy : At 4 °C water has the smallest volume.
- Medium : Ice floats because its density is lower than that of water.
- Exam : A bimetallic strip bends on heating because the two metals have different α.
ΔL = αLΔT; γ = 3α; water is densest at 4 °C.
Using α for the volume instead of γ = 3α.
Memory clue : "Line α, Area 2α, Volume 3α : 1-2-3."
Spot the difference
| Coefficient | Symbol | Value |
|---|---|---|
| Linear | α | α |
| Area | β | 2α |
| Volume | γ | 3α |
Specific heat capacity and calorimetry
The sand of a beach becomes very hot while the sea water stays cool. Link to earlier classes : specific heat of class 9 is now used in mixing problems.
The specific heat capacity c of a substance is the heat needed to raise the temperature of 1 kg of it by 1 K. Calorimetry is measuring heat by mixing substances in an insulated container.
Rule 1: Heat and temperature change — Q = m c ΔT (Q = heat, m = mass, c = specific heat, ΔT = temperature change). For water c = 4186 J kg⁻¹ K⁻¹.
- Easy : 2 kg of water heated from 20 °C to 70 °C needs Q = 2 × 4186 × 50 = 4.19 × 10⁵ J.
- Medium : 0.5 kg of copper (c = 385) cooled from 100 °C to 20 °C gives out Q = 0.5 × 385 × 80 = 1.54 × 10⁴ J.
- Exam : A 0.2 kg aluminium block (c = 900) at 100 °C is dropped into 0.5 kg of water at 20 °C. Heat lost = heat gained : 0.2 × 900 × (100 − T) = 0.5 × 4186 × (T − 20) ⇒ T = 26.3 °C.
Q = mcΔT; in an insulated mixture heat lost equals heat gained; water has a high specific heat.
Forgetting the mass or using the initial temperature instead of the change.
Memory clue : "Q = m c ΔT : Quantity of heat is Mass, Constant, Temperature-change."
Spot the difference
| Quantity | Meaning | Unit |
|---|---|---|
| Heat capacity | heat per kelvin for the whole body | J K⁻¹ |
| Specific heat | heat per kelvin per kg | J kg⁻¹ K⁻¹ |
Change of state and latent heat
Ice at 0 °C cools a drink better than water at 0 °C because it takes in extra heat while melting. Link to earlier classes : melting and boiling of class 7 and 9 are now counted in joules.
Latent heat L is the heat needed to change the state of 1 kg of a substance at a constant temperature.
Rule 1: Latent heat — Q = m L. For ice, L_f = 3.34 × 10⁵ J kg⁻¹ (fusion); for water, L_v = 2.26 × 10⁶ J kg⁻¹ (vaporisation) at 100 °C. The temperature stays constant during the change.
- Easy : Melting 0.5 kg of ice at 0 °C needs 0.5 × 3.34 × 10⁵ = 1.67 × 10⁵ J.
- Medium : Condensing 0.1 kg of steam at 100 °C gives out 0.1 × 2.26 × 10⁶ = 2.26 × 10⁵ J.
- Exam : To turn 1 kg of ice at −10 °C into steam at 100 °C : 1 × 2100 × 10 = 21 000 J (warm ice) + 3.34 × 10⁵ J (melt) + 1 × 4186 × 100 = 4.19 × 10⁵ J (warm water) + 2.26 × 10⁶ J (boil) = 3.03 × 10⁶ J.
Q = mL; no temperature change while the state changes; steam burns more than boiling water.
Using mcΔT during a change of state.
Memory clue : "Latent = Hidden : the heat goes into the bonds, not into the thermometer."
Spot the difference
| Process | Heat formula | Temperature |
|---|---|---|
| Warming | m c ΔT | changes |
| Melting / boiling | m L | constant |
Heat transfer : conduction and convection
A metal spoon in hot tea becomes hot, but a wooden spoon does not; a room heater warms the air near the ceiling first. Link to earlier classes : the three ways of heat transfer from class 7.
Conduction is the flow of heat through a material without any movement of the material itself. Convection is heat carried by the movement of a fluid.
Rule 1: Conduction — H = K A (T₁ − T₂) ÷ L (H = rate of heat flow, K = thermal conductivity, A = area, L = length or thickness, T₁ − T₂ = temperature difference).
- Easy : A copper rod (K = 385 W m⁻¹ K⁻¹), A = 1 × 10⁻⁴ m², L = 0.5 m, ΔT = 100 K : H = 385 × 10⁻⁴ × 100 ÷ 0.5 = 7.7 W.
- Medium : Doubling the length halves the heat flow (H ∝ 1 ÷ L).
- Exam : A window pane (K = 1.0 W m⁻¹ K⁻¹), 2 m² and 5 mm thick, with a 10 K difference between faces, conducts H = 1.0 × 2 × 10 ÷ 0.005 = 4000 W.
Rule 2: Convection — A heated fluid expands, becomes lighter and rises, while the cooler denser fluid sinks. This sets up a convection current (sea breeze, land breeze, boiling water).
- Easy : A heater is placed near the floor to warm the whole room.
- Medium : Sea breeze blows in the day, as air over the land rises.
- Exam : Convection cannot occur in a solid or in a vacuum.
H = KA ΔT ÷ L; metals have a large K; convection needs a fluid.
Mixing up K (conductivity) and k (spring constant or Boltzmann constant).
Memory clue : "Conduction : touch the hot end. Convection : the fluid carries it."
Spot the difference
| Process | Medium | Particles move along? |
|---|---|---|
| Conduction | solids mostly | no |
| Convection | fluids | yes |
Radiation, Stefan's law and Wien's law
We feel the warmth of the Sun, though nothing material connects us to it. Link to earlier classes : radiation as the third way of heat transfer.
Thermal radiation is energy sent out as electromagnetic waves by every body above absolute zero; it needs no medium. A black body absorbs all radiation falling on it.
Rule 1: Stefan–Boltzmann law — P = σ A T⁴ for a black body (σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴). For a surface at T in surroundings at T₀ : P = σ A (T⁴ − T₀⁴).
- Easy : A black body of A = 1 m² at 300 K radiates P = 5.67 × 10⁻⁸ × 300⁴ = 459 W.
- Medium : Doubling the temperature increases the radiated power 16 times.
- Exam : A 1 m² black body at 400 K in surroundings at 300 K loses a net P = 5.67 × 10⁻⁸ × (400⁴ − 300⁴) = 992 W.
Rule 2: Wien's displacement law — λ_m T = 2.9 × 10⁻³ m K (λ_m = wavelength of maximum emission). A hotter body radiates at a shorter wavelength.
- Easy : For the Sun (T = 5800 K), λ_m = 2.9 × 10⁻³ ÷ 5800 = 5.0 × 10⁻⁷ m = 500 nm, in the visible range.
- Medium : A body at 290 K peaks near 10 μm, in the infrared.
- Exam : A star twice as hot peaks at half the wavelength.
P ∝ T⁴; λ_m ∝ 1 ÷ T; use kelvin.
Using °C in P = σAT⁴.
Memory clue : "Stefan = Fourth power; Wien = Why shorter when hotter."
Spot the difference
| Law | States | Depends on |
|---|---|---|
| Stefan | power radiated | T⁴ |
| Wien | wavelength of the peak | 1 ÷ T |
Newton's law of cooling
A cup of tea cools quickly at first and slowly later. Link to earlier classes : cooling of a hot body in a cooler room.
Newton's law of cooling : the rate of loss of heat of a body is proportional to the difference between the temperature of the body and that of its surroundings (for a small difference).
Rule 1: Law of cooling — dT ÷ dt = −k (T − T₀), so T − T₀ = (T_i − T₀) e^(−kt).
- Easy : A body cools from 80 °C to 50 °C in 5 min in a room at 20 °C ⇒ ln(60 ÷ 30) = k × 5 ⇒ k = 0.139 min⁻¹.
- Medium : With the same k, cooling from 60 °C to 30 °C needs t = ln(40 ÷ 10) ÷ k = 1.386 ÷ 0.139 = 10 min.
- Exam : A hotter body loses heat faster : the rate at 80 °C (60 K above the room) is twice the rate at 50 °C (30 K above).
Cooling rate ∝ (T − T₀); the graph of T against t is an exponential curve.
Taking the cooling rate to be constant.
Memory clue : "Hot cools fast, warm cools slow — rate follows the gap."
Spot the difference
| Time | Temperature gap | Cooling rate |
|---|---|---|
| early | large | fast |
| later | small | slow |