Module 3: Waves and Thermodynamics — Worked Solutions (Preliminary Physics)
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Worked examples for Preliminary Physics Module 3: Waves and Thermodynamics. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
In short: full step-by-step worked solutions for Preliminary Physics — Module 3: Waves and Thermodynamics. Every question is worked through with the method and reasoning shown, so you can check how to get the answer, not just the final result.
How to use these
Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
For waves, the relationship $v = f\lambda$ ties speed, frequency and wavelength together. For heating, $Q = mc\Delta T$ links energy, mass, material and temperature change.
Example 1 — The wave equation for a sound wave
Question
A sound wave in air has a frequency of $680\ \text{Hz}$ and travels at $340\ \text{m s}^{-1}$. Find its wavelength, and then find the period of the wave.
Solution
Use the wave equation $v = f\lambda$.
$\lambda = \dfrac{v}{f} = \dfrac{340}{680} = 0.50\ \text{m}$.
The period is the reciprocal of frequency: $T = \dfrac{1}{f} = \dfrac{1}{680} = 1.47 \times 10^{-3}\ \text{s} \approx 1.5\ \text{ms}$.
Wavelength $0.50\ \text{m}$, period $\approx 1.5\ \text{ms}$. Keep the units straight — wavelength is a distance, period is a time.
The wave equation $v = f\lambda$ connects how fast a wave moves to how often it oscillates and how long each wave is. We want the wavelength, so rearrange to $\lambda = \dfrac{v}{f}$:
$\lambda = \dfrac{340}{680} = 0.50\ \text{m}$.
The period is the time for one full oscillation, and frequency counts oscillations per second, so they're reciprocals: $T = \dfrac{1}{f} = \dfrac{1}{680} = 1.47 \times 10^{-3}\ \text{s}$, which is about $1.5\ \text{ms}$.
So each wave is half a metre long and takes around one and a half milliseconds to pass — that reciprocal relationship between period and frequency is worth memorising because it comes up constantly.
Wave equation: $v = f\lambda$.
- $\lambda = v/f = 340/680 = 0.50\ \text{m}$
- $T = 1/f = 1/680 = 1.47 \times 10^{-3}\ \text{s} \approx 1.5\ \text{ms}$
$\lambda = 0.50\ \text{m}$; $T \approx 1.5\ \text{ms}$.
Where the marks go
- 1 mark: Selects and rearranges $v = f\lambda$ to $\lambda = v/f$
- 1 mark: Correct wavelength $0.50\ \text{m}$
- 1 mark: Correct period $T = 1/f \approx 1.5\ \text{ms}$
Key idea
The wave equation $v = f\lambda$ links speed, frequency and wavelength; period and frequency are reciprocals, $T = 1/f$.
Example 2 — Specific heat capacity and heat transfer
Question
A $0.50\ \text{kg}$ block of aluminium (specific heat capacity $900\ \text{J kg}^{-1}\,\text{K}^{-1}$) is heated from $20^\circ\text{C}$ to $80^\circ\text{C}$. Find the heat energy required. If this same amount of energy were instead supplied to $0.50\ \text{kg}$ of water (specific heat capacity $4200\ \text{J kg}^{-1}\,\text{K}^{-1}$) starting at $20^\circ\text{C}$, what would the water's final temperature be?
Solution
Heat absorbed: $Q = mc\Delta T$.
For the aluminium: $Q = 0.50 \times 900 \times (80 - 20) = 0.50 \times 900 \times 60 = 27\,000\ \text{J}$.
Now feed that $27\,000\ \text{J}$ into the water and solve for $\Delta T$: $\Delta T = \dfrac{Q}{mc} = \dfrac{27\,000}{0.50 \times 4200} = \dfrac{27\,000}{2100} = 12.857\ \text{K} \approx 12.9^\circ\text{C}$.
Final water temperature: $20 + 12.9 = 32.9^\circ\text{C}$.
Water's high specific heat is why it barely warms compared with the metal — same energy, far smaller temperature rise.
The amount of heat needed to change a substance's temperature is $Q = mc\Delta T$, where $c$ tells you how "stubborn" the material is about heating up.
For the aluminium, $\Delta T = 80 - 20 = 60\ \text{K}$ (a Celsius change equals a Kelvin change), so:
$Q = 0.50 \times 900 \times 60 = 27\,000\ \text{J}$.
Now we give that same $27\,000\ \text{J}$ to the water. Rearranging for the temperature change, $\Delta T = \dfrac{Q}{mc} = \dfrac{27\,000}{0.50 \times 4200} = \dfrac{27\,000}{2100} = 12.9\ \text{K}$.
Starting at $20^\circ\text{C}$, the water reaches $20 + 12.9 = 32.9^\circ\text{C}$. Notice the water rises by only about $13^\circ$ versus the aluminium's $60^\circ$ — because water's specific heat is far larger, it soaks up energy with little temperature change, which is exactly why it's used as a coolant.
$Q = mc\Delta T$.
- Aluminium: $Q = 0.50 \times 900 \times 60 = 27\,000\ \text{J}$
- Water: $\Delta T = Q/(mc) = 27\,000/(0.50 \times 4200) = 12.9\ \text{K}$
- Final temp: $20 + 12.9 = 32.9^\circ\text{C}$
$Q = 27\,000\ \text{J}$; water reaches $32.9^\circ\text{C}$.
Where the marks go
- 1 mark: Selects $Q = mc\Delta T$ with $\Delta T = 60\ \text{K}$
- 1 mark: Correct heat for aluminium $Q = 27\,000\ \text{J}$
- 1 mark: Rearranges to $\Delta T = Q/(mc)$ for the water
- 1 mark: Correct final water temperature $\approx 32.9^\circ\text{C}$
Key idea
$Q = mc\Delta T$ governs heating; a larger specific heat $c$ means a smaller temperature rise for the same energy.
Frequently asked questions
Step-by-step solutions to Waves and Thermodynamics questions in Preliminary Physics, with the full method shown for each — so you can follow the reasoning, not just the final answer.
Attempt each question yourself first, then compare your method — not just your answer — against the worked solution. The aim is to learn the approach so you can handle unfamiliar questions, rather than memorising answers.
Yes — they follow the NESA syllabus for Preliminary Physics, using the methods and notation expected in exams and assessments.