The shape of the lesson
By the end of the lesson, learners can
- The heating curve tells the story: Supply energy at a steady rate and the temperature climbs on the sloping parts and holds steady on the flat parts.
- c and L, defined: The specific heat capacity c is the energy needed to raise the temperature of one kilogram by one kelvin, so Q = mcΔθ.
- Warming without a state change: Q = mcΔθ.
- Changing state at constant temperature: Q = mL.
- Scales: T / K = θ / °C + 273.15; absolute zero is 0 K = −273.15 °C.
One picture every idea returns to
To warm a substance without a state change needs Q = mcΔθ. To change its state at constant temperature needs Q = mL. Both c and L are defined per unit mass.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Warming, and changing state.. Two relations carry almost every thermal calculation: one for warming a substance, one for changing its state. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 17 min | Teach and check: The heating curve tells the story | Supply energy at a steady rate and the temperature climbs on the sloping parts and holds steady on the flat parts. Project tab 3 of the lab (Heating curve, c and L) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: c and L, defined | The specific heat capacity c is the energy needed to raise the temperature of one kilogram by one kelvin, so Q = mcΔθ. | Think-Pair-Share |
| 30 to 43 min | Check and plenary | The quick-check questions from the student lesson, answered on mini whiteboards one at a time; re-teach the one most get wrong, then a random call on the last. | Individual, then whole class |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 70 marks) from the topic practice set. Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in c and L, defined, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Think-Pair-Share
Pose the question, give thirty seconds of silent thinking, then a minute to compare with a partner before a few pairs share with the class. The silent minute is the part that cannot be cut.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
No examiner trap is recorded for this lesson. Use the quiz key below to spot the distractor most of the class picks.
Specific heat and latent heat: topic worksheet
6 questions, 70 marks. Written in the style of the papers; no past paper question is reproduced. These questions cover the whole topic, so sibling lessons share them.
A metal block of mass 0.80 kg is heated by a well-lagged electric heater of power 50 W, as shown in Fig. 2.1. In a time of 4.0 minutes the temperature of the block rises from 18 °C to 52 °C.
(a) Define the specific heat capacity of a substance. [2]
(b) Calculate the electrical energy supplied by the heater in this time. [2]
(c) Assuming all this energy is transferred to the block, calculate the specific heat capacity of the metal. [3]
(d) In practice some energy is lost to the surroundings. Explain why this makes the experimental value of the specific heat capacity larger than the true value. [2]
(e) Suggest one change to the procedure that would reduce this error. [1]
(f) The same quantity of energy from part (b) is supplied instead to 0.80 kg of water, of specific heat capacity 4200 J kg⁻¹ K⁻¹. Calculate the rise in temperature and comment on its size. [3]
This question is about changes of state. Specific heat capacity of ice = 2100 J kg⁻¹ K⁻¹, specific heat capacity of water = 4200 J kg⁻¹ K⁻¹, specific latent heat of fusion of ice = 3.3 × 10⁵ J kg⁻¹.
(a) Define the specific latent heat of fusion of a substance. [2]
(b) An electric heater of power 40 W melts 72 g of ice, already at 0 °C, in 5.0 minutes. Use this to calculate the specific latent heat of fusion of ice. [3]
(c) Explain why, while the ice is melting, energy is supplied yet the temperature does not change. [2]
(d) The specific latent heat of vaporisation of water is much greater than its specific latent heat of fusion. Suggest, in molecular terms, why this is so. [2]
(e) A mass of 0.050 kg of ice at −10 °C is heated until it becomes water at 20 °C (Fig. 3.1). Calculate the total energy required. [4]
A 2.0 kW electric heater warms 1.5 kg of water. The specific heat capacity of water is 4200 J kg⁻¹ K⁻¹.
(a) Define specific heat capacity. [2]
(b) Calculate the energy needed to raise the water from 20 °C to 100 °C. [2]
(c) Calculate the time this would take if no energy were lost. [2]
(d) In practice the heating takes longer. State and explain why. [2]
(e) Calculate the initial rate of rise of temperature of the water. [2]
(f) State one assumption made in part (e). [1]
A 0.80 kg metal block is heated by a 50 W heater for 4.0 minutes, and its temperature rises by 25 °C.
(a) Calculate the electrical energy supplied. [2]
(b) Assuming no energy is lost, calculate the specific heat capacity of the metal. [2]
(c) The accepted value is lower than this. State and explain why this method overestimates the specific heat capacity. [2]
(d) Suggest one improvement that would reduce this error. [1]
(e) Using the value from (b), calculate the initial rate of rise of temperature. [2]
(f) Explain why measuring the temperature rise over a short interval improves the accuracy. [2]
A 0.30 kg block of ice at 0 °C is heated until it becomes water at 40 °C. The specific latent heat of fusion of ice is 3.3 × 10⁵ J kg⁻¹ and the specific heat capacity of water is 4200 J kg⁻¹ K⁻¹. Fig. 9.1 shows how the temperature varies with energy supplied.
(a) Define the specific latent heat of fusion. [2]
(b) Calculate the energy needed to melt the ice. [2]
(c) Calculate the energy needed to warm the resulting water from 0 °C to 40 °C. [2]
(d) Calculate the total energy supplied. [1]
(e) Explain why the temperature stays constant while the ice melts, even though energy is supplied. [2]
(f) State why the specific latent heat of vaporisation is much larger than that of fusion. [2]
A 0.50 kg copper block at 200 °C is dropped into 0.40 kg of water at 20 °C in an insulated container. This question links thermal equilibrium with specific heat capacity. Take c(copper) = 390 J kg⁻¹ K⁻¹ and c(water) = 4200 J kg⁻¹ K⁻¹.
(a) State the principle used to find the final temperature. [2]
(b) Write the energy-balance equation for the block and the water. [2]
(c) Calculate the final equilibrium temperature. [3]
(d) State the assumption made in this calculation. [1]
(e) Explain, in terms of thermal equilibrium, why the block and water end at the same temperature. [2]
(f) State whether the block and the water hold equal internal energies at equilibrium. [1]
Answer key and marking notes
- (a)The energy required per unit mass ✓ to raise the temperature of the substance by one kelvin (one degree) ✓
- (b)E = Pt = 50 × (4.0 × 60) = 50 × 240 ✓ = 1.2 × 10⁴ J ✓
- (c)Δθ = 52 − 18 = 34 K ✓; c = E / (mΔθ) = 1.2 × 10⁴ / (0.80 × 34) ✓ = 4.4 × 10² J kg⁻¹ K⁻¹ ✓
- (d)Part of the supplied energy is lost to the surroundings rather than heating the block ✓; the calculation credits all of it to the block, so the energy assigned per degree of rise is too large, giving c too high ✓
- (e)Improve the lagging / insulation (or apply a cooling correction, or use a lower power over a longer time) ✓
- (f)Δθ = E / (mc) = 1.2 × 10⁴ / (0.80 × 4200) ✓ = 3.6 K ✓; the rise is much smaller because water has a much greater specific heat capacity than the metal ✓
- (a)The energy required per unit mass ✓ to change the state from solid to liquid without any change of temperature ✓
- (b)E = Pt = 40 × 300 = 1.2 × 10⁴ J ✓; L = E / m = 1.2 × 10⁴ / 0.072 ✓ = 1.7 × 10⁵ J kg⁻¹ ✓
- (c)The energy is used to break or loosen the bonds between molecules, increasing their potential energy ✓; the mean kinetic energy, and so the temperature, does not change until melting is complete ✓
- (d)In vaporisation the molecules are separated completely to large distances against the intermolecular forces (and work is done pushing back the atmosphere) ✓, whereas fusion only loosens the lattice, so vaporisation needs much more energy ✓
- (e)Warm ice: mcΔθ = 0.050 × 2100 × 10 = 1050 J ✓; melt: mL = 0.050 × 3.3 × 10⁵ = 16500 J ✓; warm water: 0.050 × 4200 × 20 = 4200 J ✓; total = 1050 + 16500 + 4200 = 2.2 × 10⁴ J ✓
- (a)The energy required to raise the temperature of unit mass of a substance by one kelvin (or one degree Celsius) ✓✓
- (b)Q = mcΔθ = 1.5 × 4200 × 80 ✓ = 5.0 × 10⁵ J ✓
- (c)t = Q/P = 504000 / 2000 ✓ = 2.5 × 10² s ✓
- (d)Energy is lost to the container and surroundings ✓, so more than 504 kJ must be supplied, taking longer ✓
- (e)dθ/dt = P/(mc) = 2000 / (1.5 × 4200) ✓ = 0.32 K s⁻¹ ✓
- (f)All the power heats the water, with no losses ✓
- (a)E = Pt = 50 × (4.0 × 60) ✓ = 1.2 × 10⁴ J ✓
- (b)c = E/(mΔθ) = 12000 / (0.80 × 25) ✓ = 6.0 × 10² J kg⁻¹ K⁻¹ ✓
- (c)Some energy is lost to the surroundings, but all of it is attributed to the block ✓, so the calculated c is larger than the true value ✓
- (d)Insulate (lag) the block, or correct for the heat loss ✓
- (e)dθ/dt = P/(mc) = 50 / (0.80 × 600) ✓ = 0.10 K s⁻¹ ✓
- (f)Less time means less energy is lost to the surroundings, so the result is closer to ideal ✓✓
- (a)The energy required to change unit mass of a substance from solid to liquid with no change in temperature ✓✓
- (b)Q = mL = 0.30 × 3.3 × 10⁵ ✓ = 9.9 × 10⁴ J ✓
- (c)Q = mcΔθ = 0.30 × 4200 × 40 ✓ = 5.0 × 10⁴ J ✓
- (d)Total = 99000 + 50400 = 1.5 × 10⁵ J ✓
- (e)The energy separates molecules (increases their potential energy) rather than their kinetic energy ✓, so the temperature does not change ✓
- (f)Vaporisation fully separates the molecules and does work pushing back the atmosphere ✓, a far larger change than the small loosening during melting ✓
- (a)Conservation of energy: the thermal energy lost by the copper equals that gained by the water ✓✓
- (b)m(Cu)c(Cu)(200 − T) = m(w)c(w)(T − 20) ✓✓
- (c)0.50 × 390 × (200 − T) = 0.40 × 4200 × (T − 20) ✓; 195(200 − T) = 1680(T − 20) ✓; T = 39 °C ✓
- (d)No energy is lost to the container or surroundings ✓
- (e)Energy flows from the hotter copper to the cooler water until their temperatures are equal, when there is no net flow ✓✓
- (f)No: they share a temperature, not an internal energy ✓
The lesson's own checks as a quiz
The quick-check and exam-style questions from the student lesson, printable with the key.
The specific heat capacity of a substance is the energy needed to:
- Araise the temperature of 1 kg of it by 1 K
- Braise the temperature of any mass by 1 K
- Cmelt 1 kg of it
- Draise its temperature to 1 K
During a change of state the temperature stays constant because the supplied energy:
- Abreaks or loosens bonds, raising potential energy, not kinetic energy
- Bis reflected back out
- Craises the kinetic energy of the particles
- Dis lost to the surroundings
2.0 kg of water, of specific heat capacity 4200 J kg⁻¹ K⁻¹, is warmed by 5.0 K. The energy needed Q = mcΔθ is:
- A4.2 × 10⁴ J
- B2.1 × 10⁴ J
- C8.4 × 10⁴ J
- D4.2 × 10³ J
Compared with the specific latent heat of fusion, the specific latent heat of vaporisation of a substance is usually:
- Amuch greater
- Bmuch smaller
- Cexactly equal
- Dzero
Quiz key
Specific heat capacity is per unit mass per unit temperature rise: Q = mcΔθ.
At a state change the energy increases the molecular potential energy, so the mean kinetic energy, and thus the temperature, does not change.
Q = mcΔθ = 2.0 × 4200 × 5.0 = 42000 J = 4.2 × 10⁴ J.
Vaporisation separates molecules completely against the intermolecular forces, needing far more energy than merely melting the lattice.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.