Physics · Lesson 32

Practical 5: specific heat, latent heat and the resonance tube

WAECNECO

What the syllabus asks

WAEC: specific heat capacity by the method of mixtures, specific latent heat of fusion of ice, cooling curves; resonance tube (speed of sound in air, end correction). Links to p-units on heat and waves.

NECO: specific heat capacity by the method of mixtures, specific latent heat of fusion of ice, cooling curves; resonance tube (speed of sound in air, end correction). Links to p-units on heat and waves.

What you will learn

Determine the specific heat capacity of a metal by the method of mixtures.
Use a graph to find c without knowing the calorimeter’s heat capacity.
Determine the specific latent heat of fusion of ice.
Plot and interpret cooling curves using tangents.
Determine the speed of sound in air using a resonance tube.
Find the end correction from a graph.
State precautions and sources of error for heat and sound experiments.
Answer part (b) items on heat capacity, latent heat and resonance.

Key points

Heat lost = heat gained (no losses).
Read the boiling-water temperature; transfer quickly; highest steady θ.
m against ratio: c = slope × c_w/M; sample c = 899 J kg⁻¹ K⁻¹.
Ice: dried, melting; balance above/below room; sample L = 3.35 × 10⁵ J kg⁻¹.
Cooling curve: smooth curve; rate from a tangent.
Resonance: l = (v/4)(1/f) − e.
Sample: slope 8710 cm s⁻¹, v = 348 m s⁻¹, e = 1.2 cm.
l₂ − l₁ = λ/2 removes the end correction.
Strike forks on a rubber pad; loudest sound.
Convert grams to kilograms.

Start here: Concept summary: heat and sound in the practical

Heat experiments in WAEC Paper 3 and NECO Paper I almost always use the method of mixtures: a hot body is put into a cold liquid (or a cold body into a warm liquid) in a lagged calorimeter, and the final steady temperature is read. Sound experiments use the resonance tube: a tube closed at the bottom by water, whose air column is adjusted until it booms loudly with a tuning fork. Both rest on simple principles, and both reward careful technique more than clever calculation.

The principle of mixtures is conservation of energy: provided no heat is lost to the surroundings,

heat lost by the hot body = heat gained by the cold bodies

and the heat for a temperature change is Q = mcΔθ, where c is the specific heat capacity. For a change of state at constant temperature, Q = mL, where L is the specific latent heat.

The resonance tube rests on standing waves in a closed pipe: the water surface is a node and the open end is (approximately) an antinode. At the first resonance the air column is about a quarter of a wavelength long. The antinode actually forms slightly above the open end, by an amount called the end correction e, so l + e = λ/4. With v = fλ this leads to the graph method of section 32.8.

Precise data values used here (and accepted in WAEC): specific heat capacity of water 4200 J kg⁻¹ K⁻¹; of copper about 390 to 400 J kg⁻¹ K⁻¹; of aluminium about 900 J kg⁻¹ K⁻¹; specific latent heat of fusion of ice about 3.34 × 10⁵ J kg⁻¹ (often rounded to 3.36 × 10⁵ in examinations); speed of sound in air about 330 m s⁻¹ at 0 °C, rising by about 0.6 m s⁻¹ per °C, so about 347 m s⁻¹ at 28 °C. Use whichever values a question gives you.

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Practice questions

1. A 0.150 kg copper block at 100 °C is put in 0.100 kg of water at 27.0 °C in a calorimeter of heat capacity 20 J K⁻¹. The final temperature is 35.5 °C. Find c of copper.

Show worked answer

Heat gained = (0.100 × 4200 + 20)(8.5) = 440 × 8.5 = 3740 J. Heat lost = 0.150 × c × 64.5 = 9.675c. c = 3740/9.675 = 387 J kg⁻¹ K⁻¹.

2. State three precautions in the specific heat experiment.

Show worked answer

Lag and cover the calorimeter; transfer the block quickly; stir and record the highest steady temperature; measure the boiling water temperature.

3. Why does heat loss make the calculated c too small?

Show worked answer

The water reaches a lower final temperature than it should, so less heat appears to have been gained; dividing this smaller heat by M(θ₁ − θ), which is larger, gives a smaller c.

4. Water 0.120 kg at 35.0 °C in a calorimeter (C = 18 J K⁻¹) cools to 15.0 °C when 0.0200 kg of dried ice is added. Find L.

Show worked answer

Heat lost = (504 + 18)(20.0) = 10 440 J. Melted ice warming = 0.0200 × 4200 × 15.0 = 1260 J. L = (10 440 − 1260)/0.0200 = 4.59 × 10⁵ J kg⁻¹ (a poor result: heat gained from the room, or wet ice, would be suspected).

5. Why should ice be at its melting point when added?

Show worked answer

So that its temperature is known to be 0 °C; colder ice would also need heat to warm it to 0 °C, which the equation ignores.

Key words

Specific heat capacity: Heat to raise 1 kg by 1 K.
Heat capacity: Heat to raise a whole body by 1 K.
Method of mixtures: Finding c or L from heat exchange in a calorimeter.
Calorimeter: Lagged metal vessel used in heat experiments.
Specific latent heat of fusion: Heat to melt 1 kg at the melting point.
Cooling curve: Graph of temperature against time as a body cools.
Tangent: Line touching a curve at one point.
Resonance: Large vibration when driven at the natural frequency.
Node: Point of zero displacement in a standing wave.
Antinode: Point of maximum displacement.
End correction: Distance of the antinode above the open end.
Closed pipe: Pipe closed at one end; odd harmonics only.

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