Level 4 — ApplicationThermodynamics

Thermodynamics

⏱ 60 minutes50 marksprintable — key stays hidden on paper

Level 4: Application (Novel Problems, No Hints)

Time limit: 60 minutes Total marks: 50 Instructions: Answer all questions. Use $R = 8.314\ \mathrm{J\,mol^{-1}K^{-1}}$ , $\sigma = 5.67\times10^{-8}\ \mathrm{W\,m^{-2}K^{-4}}$ , $k_B = 1.38\times10^{-23}\ \mathrm{J/K}$ , $N_A = 6.022\times10^{23}\ \mathrm{mol^{-1}}$ . Show all working.

Question 1 — Calorimetry with phase change (10 marks)

A sealed copper vessel of mass $0.150\ \mathrm{kg}$ contains $0.200\ \mathrm{kg}$ of water, all at $18.0^\circ\mathrm{C}$ . A block of ice of mass $0.0800\ \mathrm{kg}$ at $-12.0^\circ\mathrm{C}$ is dropped in and the system sealed.

Data: $c_\text{Cu}=386$ , $c_\text{water}=4186$ , $c_\text{ice}=2100$ (all $\mathrm{J\,kg^{-1}K^{-1}}$ ), latent heat of fusion $L_f=3.34\times10^{5}\ \mathrm{J/kg}$ .

(a) Determine the final equilibrium temperature of the system, showing that not all the ice necessarily melts (or that it does). (7) (b) State the final composition (masses of ice and liquid water present). (3)

Question 2 — Adiabatic + heat transfer chain (12 marks)

$0.400\ \mathrm{mol}$ of an ideal diatomic gas ( $\gamma = 7/5$ ) initially at $300\ \mathrm{K}$ and volume $8.00\times10^{-3}\ \mathrm{m^3}$ is compressed adiabatically and reversibly to one quarter of its volume.

(a) Find the final temperature and pressure. (4) (b) Find the work done on the gas during this compression. (3) (c) The gas (now at the compressed state) is placed in contact with a reservoir and allowed to cool at constant volume back to $300\ \mathrm{K}$ . Find the heat expelled in this step. (3) (d) The compressed hot gas, before cooling, is contained by a spherical steel shell of surface area $0.0300\ \mathrm{m^2}$ and emissivity $0.60$ . Treating it as a grey body radiating into surroundings at $300\ \mathrm{K}$ , find the net instantaneous radiated power at the instant compression ends. (2)

Question 3 — Kinetic theory / Maxwell–Boltzmann (10 marks)

(a) For a gas of molecular mass $m$ at temperature $T$ , starting from the Maxwell–Boltzmann speed distribution $f(v) = 4\pi n\left(\frac{m}{2\pi k_B T}\right)^{3/2} v^2 e^{-mv^2/2k_B T},$ derive expressions for the most probable speed $v_p$ and show that $v_{rms}/v_p = \sqrt{3/2}$ . (5)

(b) Nitrogen gas ( $M = 28.0\ \mathrm{g/mol}$ ) is at $350\ \mathrm{K}$ . Compute $v_p$ , the mean speed $\bar v$ , and $v_{rms}$ numerically. (3)

(c) A rocket propulsion designer wants the fraction of molecules with speed above some threshold. Without full integration, explain qualitatively (using the distribution) why heating the gas from $350\ \mathrm{K}$ to $700\ \mathrm{K}$ increases the high-speed tail population more than proportionally, and what happens to the peak height of $f(v)$ . (2)

Question 4 — Cycle efficiency and entropy (12 marks)

An inventor claims a heat engine operating between reservoirs at $T_H = 620\ \mathrm{K}$ and $T_C = 310\ \mathrm{K}$ that absorbs $Q_H = 1000\ \mathrm{J}$ per cycle and delivers $560\ \mathrm{J}$ of work.

(a) Compute the claimed efficiency and the Carnot efficiency; state whether the claim is possible and justify using the second law. (4)

(b) For the claimed engine, compute the total entropy change of the universe per cycle ( $\Delta S_{univ}$ ) and comment. (4)

(c) A valid engine on the same reservoirs is instead run as a Carnot engine. It is then reversed to act as a refrigerator extracting heat from the $310\ \mathrm{K}$ space. Compute the coefficient of performance (COP) of this refrigerator. (2)

(d) If the refrigerator in (c) must extract $2000\ \mathrm{J}$ from the cold space per cycle, find the work input required. (2)

Question 5 — Thermal expansion (short applied) (6 marks)

A steel measuring tape is calibrated correct at $20^\circ\mathrm{C}$ ( $\alpha_\text{steel} = 1.2\times10^{-5}\ \mathrm{K^{-1}}$ ). On a hot day at $42^\circ\mathrm{C}$ a surveyor reads a distance as $86.400\ \mathrm{m}$ using this tape.

(a) Is the true distance greater or less than the reading? Explain in one sentence. (2) (b) Compute the true distance. (4)

Answer keyMark scheme & solutions

Question 1 (10 marks)

Strategy: Compare heat available from cooling water+vessel to $0^\circ$ C against heat needed to warm ice to $0^\circ$ C and melt it.

(a) Heat released by water + copper cooling from $18\to0^\circ$ C: $Q_\text{avail} = (m_wc_w + m_{Cu}c_{Cu})(18.0) = (0.200\cdot4186 + 0.150\cdot386)(18.0)$ $= (837.2 + 57.9)(18.0) = 895.1\times18.0 = 1.6112\times10^4\ \mathrm{J}$ (2)

Heat to warm ice from $-12\to0^\circ$ C: $Q_\text{warm} = m_i c_\text{ice}(12.0) = 0.0800\cdot2100\cdot12.0 = 2016\ \mathrm{J}$ (1)

Heat to melt all ice: $Q_\text{melt,all} = m_i L_f = 0.0800\cdot3.34\times10^5 = 2.672\times10^4\ \mathrm{J}$ (1)

Remaining after warming ice: $1.6112\times10^4 - 2016 = 1.4096\times10^4\ \mathrm{J}$ . Since $1.4096\times10^4 < 2.672\times10^4$ , not all ice melts — system settles at $0^\circ$ C. (2)

Final temperature $= 0^\circ\mathrm{C}$ (1)

(b) Mass of ice melted: $m_\text{melt} = \frac{1.4096\times10^4}{3.34\times10^5} = 0.0422\ \mathrm{kg}$ Ice remaining $= 0.0800 - 0.0422 = 0.0378\ \mathrm{kg}$ . Liquid water $= 0.200 + 0.0422 = 0.2422\ \mathrm{kg}$ . (3) (Composition: ≈0.0378 kg ice + ≈0.242 kg water at 0°C.)

Question 2 (12 marks)

(a) Adiabatic: $TV^{\gamma-1}=$ const, $\gamma-1=0.4$ , $V_f/V_i=1/4$ . $T_f = T_i(V_i/V_f)^{\gamma-1} = 300\cdot4^{0.4} = 300\cdot1.7411 = 522.3\ \mathrm{K}$ (2) Initial pressure $P_i = nRT_i/V_i = 0.400\cdot8.314\cdot300/8.00\times10^{-3} = 1.2471\times10^5\ \mathrm{Pa}$ . $P_f = P_i(V_i/V_f)^{\gamma} = 1.2471\times10^5\cdot4^{1.4} = 1.2471\times10^5\cdot6.9644 = 8.686\times10^5\ \mathrm{Pa}$ (2)

(b) Adiabatic $Q=0$ , so work on gas $= \Delta U = nC_V\Delta T$ , $C_V=\tfrac52R$ . $W_\text{on} = nC_V(T_f-T_i) = 0.400\cdot\tfrac52\cdot8.314\cdot(522.3-300)$ $= 0.400\cdot20.785\cdot222.3 = 1848\ \mathrm{J}$ (3)

(c) Constant-volume cooling $522.3\to300$ K: $Q = nC_V\Delta T = 0.400\cdot20.785\cdot(300-522.3) = -1848\ \mathrm{J}$ Heat expelled $\approx 1848\ \mathrm{J}$ (magnitude). (3)

(d) Net radiated power: $P = \varepsilon\sigma A(T_f^4 - T_s^4) = 0.60\cdot5.67\times10^{-8}\cdot0.0300\cdot(522.3^4 - 300^4)$ $522.3^4 = 7.442\times10^{10}$ , $300^4=8.10\times10^{9}$ , difference $=6.632\times10^{10}$ . $P = 0.60\cdot5.67\times10^{-8}\cdot0.0300\cdot6.632\times10^{10} = 67.7\ \mathrm{W}$ (2)

Question 3 (10 marks)

(a) $v_p$ : maximise $f(v)$ ; set $\frac{d}{dv}(v^2 e^{-mv^2/2k_BT})=0$ : $2v e^{-\cdots} + v^2(-mv/k_BT)e^{-\cdots}=0 \Rightarrow 2 = \frac{mv^2}{k_BT} \Rightarrow v_p=\sqrt{\frac{2k_BT}{m}}$ (2) $v_{rms}=\sqrt{\langle v^2\rangle}=\sqrt{3k_BT/m}$ (standard result / from $\int v^2 f\,dv$ ). (2) $\frac{v_{rms}}{v_p}=\sqrt{\frac{3k_BT/m}{2k_BT/m}}=\sqrt{3/2}.$ (1)

(b) $M=0.0280\ \mathrm{kg/mol}$ ; use $k_BT/m = RT/M$ , $RT/M = 8.314\cdot350/0.0280 = 1.0393\times10^5$ . $v_p=\sqrt{2\cdot1.0393\times10^5}=456\ \mathrm{m/s}$ $\bar v=\sqrt{\frac{8k_BT}{\pi m}}=\sqrt{8/\pi}\,\sqrt{RT/M}=1.5958\cdot322.4=514\ \mathrm{m/s}$ $v_{rms}=\sqrt{3\cdot1.0393\times10^5}=558\ \mathrm{m/s}$ (3)

(c) Raising $T$ shifts $v_p\propto\sqrt T$ to higher speed and broadens the distribution; the exponential $e^{-mv^2/2k_BT}$ decays more slowly, so the tail beyond a fixed threshold grows disproportionately (exponentially sensitive). Normalisation (area fixed) plus broadening means the peak height decreases. (2)

Question 4 (12 marks)

(a) Claimed $\eta = 560/1000 = 0.560$ (56.0%). Carnot $\eta_C = 1 - T_C/T_H = 1 - 310/620 = 0.500$ (50.0%). (2) Since claimed $\eta > \eta_C$ , the engine violates the second law (Kelvin–Planck / Carnot's theorem: no engine can exceed Carnot efficiency between the same reservoirs). Impossible. (2)

(b) For claimed engine, $Q_C = Q_H - W = 1000-560 = 440\ \mathrm{J}$ dumped to cold reservoir. $\Delta S_{univ}=-\frac{Q_H}{T_H}+\frac{Q_C}{T_C}=-\frac{1000}{620}+\frac{440}{310}=-1.6129+1.4194=-0.1935\ \mathrm{J/K}$ (3) $\Delta S_{univ}<0$ violates the second law (entropy of universe must not decrease) — confirms impossibility. (1)

(c) Reversed Carnot refrigerator: $\mathrm{COP}=\dfrac{T_C}{T_H-T_C}=\dfrac{310}{620-310}=\dfrac{310}{310}=1.00$ . (2)

(d) $\mathrm{COP}=Q_C/W \Rightarrow W = Q_C/\mathrm{COP} = 2000/1.00 = 2000\ \mathrm{J}$ . (2)

Question 5 (6 marks)

(a) At $42^\circ$ C the steel tape has expanded, so each marked interval is longer than nominal; the tape under-reads, hence the true distance is greater than the reading. (2)

(b) $\Delta T = 22\ \mathrm{K}$ . Each division length scales by $(1+\alpha\Delta T)$ : $L_\text{true}=L_\text{read}(1+\alpha\Delta T)=86.400(1+1.2\times10^{-5}\cdot22)$ $=86.400(1+2.64\times10^{-4})=86.400+0.02281 = 86.423\ \mathrm{m}$ (4)

[
  {"claim":"Q1: heat available (16112 J) < warm+melt-all (2016+26720), and melted ice mass ~0.0422 kg",
   "code":"Qav=(0.200*4186+0.150*386)*18.0; Qwarm=0.0800*2100*12.0; Qmeltall=0.0800*3.34e5; mmelt=(Qav-Qwarm)/3.34e5; result = (Qav < Qwarm+Qmeltall) and abs(mmelt-0.0422)<0.001"},
  {"claim":"Q2a: adiabatic final T = 300*4**0.4 ≈ 522.3 K",
   "code":"Tf=300*4**0.4; result = abs(Tf-522.3)<0.5"},
  {"claim":"Q2b: work on gas ≈1848 J and equals Q2c magnitude",
   "code":"Tf=300*4**0.4; W=0.400*2.5*8.314*(Tf-300); result = abs(W-1848)<3"},
  {"claim":"Q4a: claimed eff 0.56 exceeds Carnot 0.50",
   "code":"eta=560/1000; etac=1-310/620; result = (eta>etac) and abs(etac-0.5)<1e-9"},
  {"claim":"Q4b: dS_univ negative ≈ -0.1935 J/K",
   "code":"dS=-1000/620+440/310; result = (dS<0) and abs(dS+0.1935)<0.001"},
  {"claim":"Q5b: true distance ≈86.423 m",
   "code":"L=86.400*(1+1.2e-5*22); result = abs(L-86.4228)<0.001"}
]