Physics notes, chapter 4
Heat
Textbook Physics Form 4 KSSM (DLP), pages 118–168
What this chapter covers
- Explain thermal equilibrium and use it to calibrate a liquid-in-glass thermometer with two fixed points (4.1)
- Define heat capacity and specific heat capacity, and solve problems using Q = mcΔθ and Pt = mcΔθ (4.2)
- Explain applications of specific heat capacity in daily life, materials and nature (4.2)
- Explain latent heat, specific latent heat of fusion and of vaporisation, and read heating and cooling curves (4.3)
- Solve problems using Q = ml and describe applications such as the refrigerator and cooling by evaporation (4.3)
- Explain gas pressure, temperature and volume with the kinetic theory of gas (4.4)
- State and apply Boyle's, Charles' and Gay-Lussac's laws, using absolute temperature in kelvin (4.4)
4.1Thermal equilibrium
Two objects are in thermal contact when heat can flow between them. They are in thermal equilibrium when they are at the same temperature, so the net rate of heat transfer is zero.
When a hot object P touches a cold object Q, heat flows from P to Q faster than from Q to P. The temperature of P falls and the temperature of Q rises, until both are equal. After that, heat still moves in both directions but at the same rate, so there is no net flow.
Textbook figure: Physics Form 4 KSSM (DLP), Figure 4.2, p. 121.
| Situation | How thermal equilibrium is used |
|---|---|
| Clinical thermometer under the tongue | Heat flows from the body to the thermometer until both have the same temperature, so the reading is the body temperature |
| Food in a refrigerator | Heat flows from the food to the cold air until they reach the same temperature, so the food stays fresh longer |
| Ice cubes in a drink | Ice absorbs heat from the drink and melts; the drink cools until equilibrium |
Calibrating a liquid-in-glass thermometer
A thermometer with no scale is calibrated with two fixed points of distilled water: the melting point of ice, 0 °C (lower fixed point) and the boiling point of water, 100 °C (upper fixed point). The thermometric property used is the length of the liquid column, which changes with temperature.
Steps
1. Put the bulb in ice and distilled water. When the column stops moving, mark it as 0 °C.2. Put the bulb in boiling distilled water. When the column stops moving, mark it as 100 °C. Measure this length as L100.
3. For an unknown liquid, measure the length from the 0 °C mark as Lθ. Then θ = (Lθ ÷ L100) × 100 °C.
Worked example
The 0 °C and 100 °C marks are 25 cm apart. The liquid column in a bath is 10 cm long (from the 0 °C mark).θ = (10 ÷ 25) × 100 = 40 °C.
Wait until the column stops moving before marking it. At that moment the thermometer and the liquid are in thermal equilibrium. Keep the bulb off the base and sides of the beaker, which may be hotter or colder than the liquid.
Two objects at the same temperature are in thermal contact. Is there any heat transfer between them?
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Heat moves in both directions at the same rate, so the net heat transfer is zero. They are in thermal equilibrium.
4.2Specific heat capacity
Heat capacity, C is the heat needed to raise the temperature of an object by 1 °C: C = Q ÷ Δθ, unit J °C−1.
Specific heat capacity, c is the heat needed to raise the temperature of 1 kg of a substance by 1 °C: c = Q ÷ (mΔθ), unit J kg−1 °C−1 (or J kg−1 K−1).
Specific heat capacity, c is the heat needed to raise the temperature of 1 kg of a substance by 1 °C: c = Q ÷ (mΔθ), unit J kg−1 °C−1 (or J kg−1 K−1).
Heat capacity depends on the mass of the object, while specific heat capacity depends only on the material. The heat absorbed or released is Q = mcΔθ. For an electric heater of power P switched on for time t, Q = Pt, so Pt = mcΔθ.
| Substance | c / J kg−1 °C−1 | Substance | c / J kg−1 °C−1 |
|---|---|---|---|
| Water | 4 200 | Aluminium | 900 |
| Sea water | 3 900 | Iron | 450 |
| Cooking oil | 1 850 | Copper | 390 |
| Wood | 1 700 | Gold | 300 |
| Sand | 800 | Lead | 130 |
Values selected from Table 4.2, p. 128.
Mixing up C and c. Heat capacity C is for a whole object (unit J °C−1); specific heat capacity c is per kilogram (unit J kg−1 °C−1). Also convert grams to kilograms before using Q = mcΔθ.
Worked examples
Example 1: finding c with a heater
A 0.60 kg metal block is heated by an 80 W heater for 90 s. Its temperature rises from 20 °C to 50 °C.Δθ = 50 − 20 = 30 °C. c = Pt ÷ (mΔθ) = (80 × 90) ÷ (0.60 × 30) = 400 J kg−1 °C−1.
Assumption: all the heat from the heater is absorbed by the block and none is lost to the surroundings.
Example 2: mixing
20 g of boiling water (100 °C) is poured into 200 g of water at 28 °C. Let the final temperature be y.Heat released = heat absorbed: 0.02 × 4200 × (100 − y) = 0.20 × 4200 × (y − 28).
Cancel 4200: 2 − 0.02y = 0.20y − 5.6, so 7.6 = 0.22y and y = 34.5 °C.
Assumption: no heat is lost to the surroundings or the glass.
Applications
| Example | Reason |
|---|---|
| Water in a car radiator | High c: absorbs a large amount of engine heat with a small rise in temperature |
| Metal wok and aluminium pot body | Low c: heats up quickly |
| Plastic handle, clay pot, wooden houses | High c: heats up slowly, so it is safe to touch or keeps heat in or out |
| Space capsule outer layer | High specific heat capacity and high melting point resist the heat from air friction |
| Sea breeze (day) | Land has a lower c than sea, so land gets hotter; hot air above land rises and cooler sea air moves inland |
| Land breeze (night) | Land cools faster than sea; warmer air over the sea rises and cooler air from land moves out to sea |
1.5 kg of water is heated from 28 °C to 48 °C. How much heat is absorbed? (c = 4200 J kg−1 °C−1)
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Q = mcΔθ = 1.5 × 4200 × (48 − 28) = 1.5 × 4200 × 20 = 126 000 J.
4.3Specific latent heat
Latent heat is the heat absorbed or released during a change of phase without a change in temperature. It is absorbed in melting and boiling, and released in freezing and condensation.
Specific latent heat, l is the heat absorbed or released when 1 kg of a substance changes phase at constant temperature: l = Q ÷ m, so Q = ml, unit J kg−1.
Specific latent heat of fusion, lf: solid and liquid. Specific latent heat of vaporisation, lv: liquid and gas.
Specific latent heat of fusion, lf: solid and liquid. Specific latent heat of vaporisation, lv: liquid and gas.
On the flat parts of the curve the heat supplied does not raise the average kinetic energy of the molecules. It is used to weaken the bonds (melting) or break the bonds and separate the molecules far apart (boiling). So the temperature stays constant. On a cooling curve the flat parts show heat being released as bonds form again.
| Water | Value |
|---|---|
| lf of ice | 3.34 × 105 J kg−1 |
| lv of water | 2.26 × 106 J kg−1 |
For the same substance lv is larger than lf, because the molecules must be completely separated when a liquid boils.
Worked example: one change of phase
A 480 W heater melts ice for 120 s. The balance reading falls by 0.172 kg.Pt = mlf → 480 × 120 = 0.172 × lf → lf = 57 600 ÷ 0.172 = 3.35 × 105 J kg−1.
Worked example: two stages
Heat to change 0.75 kg of water at 30 °C into steam at 100 °C:Stage 1: Q1 = mcΔθ = 0.75 × 4200 × 70 = 220 500 J.
Stage 2: Q2 = mlv = 0.75 × 2.26 × 106 = 1 695 000 J.
Total Q = Q1 + Q2 = 1.92 × 106 J.
Use Q = mcΔθ when the temperature changes and Q = ml when the phase changes. If both happen, add the stages.
Applications
- Evaporation cools: liquid molecules absorb latent heat from the surroundings (for example sweat takes heat from the body; air flow speeds up evaporation).
- Ice cubes cool a drink: melting ice absorbs lf from the drink.
- Steaming food: steam condenses on the food and releases lv, which heats the food.
Textbook figure: Physics Form 4 KSSM (DLP), Figure 4.21, p. 145.
Thinking that temperature rises while a substance is melting or boiling. The temperature stays constant; all the heat goes into changing the phase.
How much heat is needed to boil away 0.2 kg of water already at 100 °C? (lv = 2.26 × 106 J kg−1)
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Q = mlv = 0.2 × 2.26 × 106 = 4.52 × 105 J.
4.4Gas laws
Gas molecules move randomly and fill their container. Pressure is the force per unit area produced when molecules collide with the walls. A higher temperature means a greater average kinetic energy. Volume of a gas is the volume of its container.
| Quantity | S.I. unit | Other units |
|---|---|---|
| Pressure, P | pascal (Pa) | cm Hg, kPa |
| Temperature, T | kelvin (K) | °C |
| Volume, V | m3 | cm3, ml |
Using °C in a gas law calculation for temperature. Always convert first: T (K) = θ (°C) + 273. Absolute zero is 0 K = −273 °C.
| Law | Statement | Constant | Formula | Kinetic theory explanation |
|---|---|---|---|---|
| Boyle | P is inversely proportional to V | Mass and temperature | P1V1 = P2V2 | Smaller volume: more molecules per unit volume, more frequent collisions, so more force per area |
| Charles | V is directly proportional to T (kelvin) | Mass and pressure | V1/T1 = V2/T2 | Hotter: faster molecules, so volume must increase to keep the collision rate and pressure unchanged |
| Gay-Lussac | P is directly proportional to T (kelvin) | Mass and volume | P1/T1 = P2/T2 | Hotter: faster molecules hit the walls more often and harder, so pressure increases |
Boyle's law
Air in a syringe: 60 cm3 at 108 kPa is compressed to 48 cm3.P2 = (108 × 60) ÷ 48 = 135 kPa.
Charles' law
An air bubble of 1.2 cm3 at 27 °C is warmed to 47 °C. T1 = 300 K, T2 = 320 K.V2 = (1.2 × 320) ÷ 300 = 1.28 cm3.
Gay-Lussac's law
Gas in a steel cylinder: 180 kPa at 25 °C is heated to 52 °C. T1 = 298 K, T2 = 325 K.P2 = (180 × 325) ÷ 298 = 196 kPa (196.3).
Direct proportion (a straight line through the origin) works only with kelvin. If the graph uses °C, the straight line meets the temperature axis at −273 °C, where the volume or pressure would reach zero (absolute zero).
A gas cylinder is at 210 kPa and 25 °C. After a journey the pressure is 240 kPa. Find the temperature. (Volume unchanged.)
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T1 = 298 K. T2 = (240 × 298) ÷ 210 = 340.6 K, so θ = 340.6 − 273 = 67.6 °C.