Worked Solutions
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.
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.