a. Find the final temperature of the two objects, and the total work done by... · OlympiadBench #1434

Physics · thermodynamics

Problem

a. Find the final temperature TfT_{f} of the two objects, and the total work WW done by the engine. Given constants: g=9.8 N/kgg=9.8 \mathrm{~N} / \mathrm{kg} k=1/4πϵ0=8.99×109 N⋅m2/C2k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} c=3.00×108 m/sc=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} NA=6.02×1023( mol)−1N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} σ=5.67×10−8 J/(s⋅m2⋅K4)\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) 1eV=1.602×10−19 J1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} me=9.109×10−31 kg=0.511MeV/c2m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} mp=1.673×10−27 kg=938MeV/c2m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2} sin⁡θ≈θ−16θ3\sin \theta \approx \theta-\frac{1}{6} \theta^{3} for ∣θ∣≪1|\theta| \ll 1 G=6.67×10−11 N⋅m2/kg2km=μ0/4π=10−7 T⋅m/AkB=1.38×10−23 J/KR=NAkB=8.31 J/(mol⋅K)e=1.602×10−19Ch=6.63×10−34 J⋅s=4.14×10−15eV⋅s(1+x)n≈1+nx for ∣x∣≪1ln⁡(1+x)≈x for ∣x∣≪1cos⁡θ≈1−12θ2 for ∣θ∣≪1\begin{aligned} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ & e=1.602 \times 10^{-19} \mathrm{C} \\ & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ & \ln (1+x) \approx x \text { for }|x| \ll 1 \\ & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{aligned}  velocity parameter β=vc Lorentz factor γ=11−β2 relativistic momentum p=γβmc relativistic energy E=γmc2 relativistic doppler shift ff0=1−β1+β\begin{array}{lrl} \text { velocity parameter } & \beta & =\frac{v}{c} \\ \text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\ \text { relativistic momentum } & p & =\gamma \beta m c \\ \text { relativistic energy } & E & =\gamma m c^{2} \\ \text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}} \end{array} Extra Supplementary Reading Materials: Consider two objects with equal heat capacities CC and initial temperatures T1T_{1} and T2T_{2}. A Carnot engine is run using these objects as its hot and cold reservoirs until they are at equal temperatures. Assume that the temperature changes of both the hot and cold reservoirs is very small compared to the temperature during any one cycle of the Carnot engine.
Intuitive explanation

This problem asks you to apply energy conservation and entropy constraints to a system where a heat engine operates between two finite thermal reservoirs until they equilibrate. The key insight is that a reversible Carnot engine maintains zero total entropy change for the universe—so while internal energy is conserved, the entropy condition dStotal=0dS_{\text{total}} = 0 uniquely determines the final state. Use energy conservation to relate heat flows to temperature changes, then invoke the reversibility condition to find TfT_f; the work follows directly from energy balance.

Intuitive explanation

This problem asks you to apply energy conservation and entropy constraints to a system where a heat engine operates between two finite thermal reservoirs until they equilibrate. The key insight is that a reversible Carnot engine maintains zero total entropy change for the universe—so while internal energy is conserved, the entropy condition dStotal=0dS_{\text{total}} = 0 uniquely determines the final state. Use energy conservation to relate heat flows to temperature changes, then invoke the reversibility condition to find TfT_f; the work follows directly from energy balance.

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