c. Find the maximum temperature to which the third object can be raised. · OlympiadBench #1436
Physics · thermodynamics
Problem
c. Find the maximum temperature TH to which the third object can be raised.
Given constants:g=9.8N/kgk=1/4πϵ0=8.99×109N⋅m2/C2c=3.00×108m/sNA=6.02×1023(mol)−1σ=5.67×10−8J/(s⋅m2⋅K4)1eV=1.602×10−19Jme=9.109×10−31kg=0.511MeV/c2mp=1.673×10−27kg=938MeV/c2sinθ≈θ−61θ3 for ∣θ∣≪1G=6.67×10−11N⋅m2/kg2km=μ0/4π=10−7T⋅m/AkB=1.38×10−23J/KR=NAkB=8.31J/(mol⋅K)e=1.602×10−19Ch=6.63×10−34J⋅s=4.14×10−15eV⋅s(1+x)n≈1+nx for ∣x∣≪1ln(1+x)≈x for ∣x∣≪1cosθ≈1−21θ2 for ∣θ∣≪1 velocity parameter Lorentz factor relativistic momentum relativistic energy relativistic doppler shift βγpEf0f=cv=1−β21=γβmc=γmc2=1+β1−β
Extra Supplementary Reading Materials:
Consider two objects with equal heat capacities C and initial temperatures T1 and T2. 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.
Context question:
a. Find the final temperature Tf of the two objects, and the total work W done by the engine.
Context answer:
Tf=T1T2, W=C[T1+T2−2T1T2]
Extra Supplementary Reading Materials:
Now consider three objects with equal and constant heat capacity at initial temperatures T1=100K,T2=300K, and T3=300K. Suppose we wish to raise the temperature of the third object.
To do this, we could run a Carnot engine between the first and second objects, extracting work W. This work can then be dissipated as heat to raise the temperature of the third object. Even better, it can be stored and used to run a Carnot engine between the first and third object in reverse, which pumps heat into the third object.
Assume that all work produced by running engines can be stored and used without dissipation.
Context question:
b. Find the minimum temperature TL to which the first object can be lowered.
Context answer:
100