IBDP Physics- B.1 Thermal energy transfers- IB Style Questions For HL Paper 2 -FA 2025
Question
A copper rod has mass \(0.400\,\mathrm{kg}\). The density of solid copper is \(8.94\times10^{3}\,\mathrm{kg\,m^{-3}}\). The molar mass (the mass of a mole) of copper is \(63.5\,\mathrm{g\,mol^{-1}}\).
Estimate
(a)(i) the mass of a copper atom;
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(ii) the volume corresponding to each atom, given by
\(\dfrac{\text{volume of copper}}{\text{number of copper atoms}}\)
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(iii) the average separation of two neighbouring copper atoms, assuming that the volume corresponding to each atom is a cube.
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(b) The specific latent heat of fusion of copper is \(206\,\mathrm{kJ\,kg^{-1}}\). Calculate the energy needed to completely melt \(0.400\,\mathrm{kg}\) of solid copper at its melting point.
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(c) Copper atoms in the solid and liquid phase coexist at the melting point. Comparing equal numbers of atoms in the solid and liquid phase, state and explain which atoms have the greater internal energy.
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(d) A copper rod is suspended at its centre, as shown. P and Q are the equilibrium positions of two copper atoms in the rod. When a hammer taps the left-hand end of the rod, the first harmonic sound standing wave is formed in the rod. The longitudinal standing wave in the rod is identical to a standing wave in a pipe with both ends open.
(i) Explain, by reference to the principle of superposition, why a standing wave is set up in the pipe.
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(ii) Draw, on the diagram, the standing wave in the rod. The width of the rod is shown larger to help with your drawing.
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(iii) The graph shows the variation with time of the displacement of the atom at P. Draw, on the same axes, the variation with time of the displacement of the atom at Q.
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(e) The copper rod is brought close to a glass tube that contains light sand. When the left-hand end of the rod is tapped with a hammer and the length of the tube is adjusted by moving the piston, a standing wave is established in the tube. Sand collects in piles as shown.
(i) Outline why the sand collects in piles.
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(ii) The following data are available:
\(d=1.8\,\mathrm{cm}\)
speed of sound in air \(=330\,\mathrm{m\,s^{-1}}\)
length of copper rod \(=34\,\mathrm{cm}\)
Determine the speed of sound in copper.
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Most-appropriate topic codes (IB Physics HL First Assessment 2025):
• Topic B.4: Thermodynamics — parts (b), (c), latent heat and internal energy changes during melting
• Topic C.2: Wave model — parts (d)(i), (d)(ii), (d)(iii), superposition, standing waves and displacement
• Topic C.4: Standing waves and resonance — parts (d)(i), (d)(ii), (e)(i), (e)(ii), standing-wave patterns and resonance
▶️ Answer/Explanation
(a)(i) Correct Answer: \( \boxed{1.1\times10^{-25}\,\mathrm{kg}} \)
One mole of copper contains \(N_{\mathrm A}=6.02\times10^{23}\) atoms.
The molar mass is \(63.5\,\mathrm{g\,mol^{-1}}=0.0635\,\mathrm{kg\,mol^{-1}}\).
Therefore, the mass of one copper atom is
\(m_{\mathrm{atom}}=\dfrac{0.0635}{6.02\times10^{23}}\)
\(m_{\mathrm{atom}}\approx\boxed{1.1\times10^{-25}\,\mathrm{kg}}\)
(a)(ii) Correct Answer: \( \boxed{1.2\times10^{-29}\,\mathrm{m^3}} \)
The volume of the copper rod is
\(V=\dfrac{m}{\rho}=\dfrac{0.400}{8.94\times10^3}\)
\(V\approx4.47\times10^{-5}\,\mathrm{m^3}\)
The number of copper atoms is
\(N=\dfrac{0.0635}{6.02\times10^{23}}\)
Using the volume per atom,
\(V_{\mathrm{atom}}=\dfrac{4.47\times10^{-5}}{0.400/(1.055\times10^{-25})}\)
or equivalently
\(V_{\mathrm{atom}}=\dfrac{0.0635}{(8.94\times10^3)(6.02\times10^{23})}\)
\(V_{\mathrm{atom}}\approx\boxed{1.2\times10^{-29}\,\mathrm{m^3}}\)
(a)(iii) Correct Answer: \( \boxed{2.3\times10^{-10}\,\mathrm{m}} \)
If the volume associated with each atom is a cube, its side length is approximately the separation between neighbouring atoms.
\(d=\sqrt[3]{V_{\mathrm{atom}}}\)
\(d=\sqrt[3]{1.18\times10^{-29}}\)
\(d\approx\boxed{2.3\times10^{-10}\,\mathrm{m}}\)
(b) Correct Answer: \( \boxed{82.4\,\mathrm{kJ}} \)
At the melting point, the energy required is related to the specific latent heat of fusion by
\(Q=mL_{\mathrm f}\)
\(Q=(0.400)(206\times10^3)\)
\(Q=\boxed{8.24\times10^4\,\mathrm{J}=82.4\,\mathrm{kJ}}\)
(c) Correct Answer:
The atoms in the liquid phase have greater internal energy.
At the melting point, the temperature of the solid and liquid is the same, so the average kinetic energy of the atoms is the same.
However, the atoms in the liquid have greater average separation and therefore greater intermolecular potential energy.
Since internal energy is the sum of kinetic and potential energies, the liquid has the greater internal energy.
(d)(i) Correct Answer:
The wave produced by the hammer travels along the rod or pipe and is reflected at the end.
The incident and reflected waves have the same frequency and travel in opposite directions. Their superposition produces regions of constructive and destructive interference, forming a standing wave.
(d)(ii) Correct Answer:
For the first harmonic in a system with both ends open, there are antinodes at both ends and a node at the centre.
Therefore, the standing-wave pattern should show two displacement antinodes at the ends of the rod and one node at its centre.
(d)(iii) Correct Answer:
Points P and Q are on opposite sides of the central node, so they oscillate in antiphase.
Therefore, the displacement-time graph for Q should have the same frequency and amplitude as P but be inverted, corresponding to a phase difference of \(\pi\).
(e)(i) Correct Answer:
The standing wave causes the air to oscillate horizontally, producing regions of large and small displacement.
The sand is displaced away from regions of large oscillation and collects near the nodes, where the displacement is small or zero.
(e)(ii) Correct Answer: \( \boxed{6.2\times10^3\,\mathrm{m\,s^{-1}}} \)
The distance between consecutive piles of sand corresponds to half a wavelength.
Therefore,
\(\dfrac{\lambda_{\mathrm{air}}}{2}=1.8\,\mathrm{cm}\)
\(\lambda_{\mathrm{air}}=3.6\times10^{-2}\,\mathrm{m}\)
The frequency of the standing wave is
\(f=\dfrac{v_{\mathrm{air}}}{\lambda_{\mathrm{air}}}\)
\(f=\dfrac{330}{3.6\times10^{-2}}\)
\(f\approx9.17\times10^3\,\mathrm{Hz}\)
For the first harmonic in the copper rod, there are antinodes at both ends, so
\(\lambda_{\mathrm{Cu}}=2L=2(0.34)=0.68\,\mathrm{m}\)
The frequency remains unchanged when the wave enters the copper rod.
Therefore,
\(v_{\mathrm{Cu}}=f\lambda_{\mathrm{Cu}}\)
\(v_{\mathrm{Cu}}=(9.17\times10^3)(0.68)\)
Thus, \( \boxed{v_{\mathrm{Cu}}\approx6.2\times10^3\,\mathrm{m\,s^{-1}}} \).