- #1
CAF123
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Homework Statement
Consider two fixed volume bricks of mass m1=2kg and m2=1kg with initial temperatures T1=400K and T2=100K. They are enclosed in a system thermally isolated from the surroundings and are made from a material with a heat capacity cv = 1kJ/kg/K.
A) In Process 1, the bricks are brought into thermal contact. What is their final temperature once they reached thermal equilibrium with each other?
B)In Process 2, the initial conditions are the same, but the maximum amount of work is extracted from the system (e.g via a heat engine). What is the final temperature in this case?
C)Find the change in entropy of processes 1 and 2.
D)Find the amount of work extracted in Process 2.
Homework Equations
Q = cvmΔT
ΔS ≥ 0 for an isolated system.
The Attempt at a Solution
A) is fine, the final temperature should be between 100 and 400K. I get 300K which is reasonable.
B)If the max amount of work is extracted, then the heat gained by brick 2, Q2= Q1 - W. I don't see how to proceed here unless I write an explicit expression for W (but that comes in D)). I did find an expression though, but when I use it, it yields a negative temperature.
C)Entropy lost by brick 1 = entropy gained by brick 2 since system thermally isolated. Hence net change in entropy in process A is zero.
D)For a thermally isolated system, ΔS ≥ 0. The maximum amount of work is extracted from a Carnot engine, which is reversible, so the equality applies. Therefore, ΔS2-ΔS1 = 0 = Q2/T2-Q1/T1=0 = (Q1-W)/T2 - Q1/T1=0. Rearrange for W gives Q1(1-T2/T1). Is that correct?
Many thanks for any guidance.