Tesla embraces the heat engine

Classical Mechanics Level 2

Tesla decides to go back to basics with their new line, the Entropica series. Leaving aside their rather successful venture into electric engines, Tesla pivots to embrace the heat engine, which uses two heat blocks of initial temperature T + T_+ , and T − T_- to drive an engine through a thermodynamic cycle (pictured above).

Concretely, each cycle drives some heat δ Q \delta Q from the hot reservoir to the cold reservoir, and the engine performs some work δ W \delta W to propel the car, such that the engine returns to its original state after each cycle. Clearly, each cycle lowers the temperature T + T_+ , and raises the temperature T − T_- , until T + = T − = T ∗ T_+=T_-=T^* , at which point the engine can do no further work.

Find T ∗ T^* (in deg Kelvin) at the point where the engine stops running.

Assumptions and Details

  • The entire process is thermodynamically reversible.
  • T + T_+ = 423 ∘ ^\circ K
  • T − T_- = 300 ∘ ^\circ K
  • Both heat reservoirs are metal blocks of heat capacity γ \gamma


The answer is 356.230262611.

Josh Silverman by Josh Silverman

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1 solution

Josh Silverman Staff
May 10, 2015

Since each heat block has heat capacity γ \displaystyle \gamma , the heat content of each block is given by Q = γ T \displaystyle Q = \gamma T .

The process is thermodynamically reversible, which means that there is no change in the total entropy of the system.

In moving heat δ Q \displaystyle \delta Q into or out of a heat reservoir, the change in entropy is given by d S = δ Q T \displaystyle dS = \frac{\delta Q}{T} , thus the change in entropy of a heat block is given by the sum ∫ δ Q T \displaystyle \int \frac{\delta Q}{T} or, using the specific heat relation for the blocks, γ ∫ d T T \displaystyle \gamma \int \frac{dT}{T} .

The hot block (initially at T + \displaystyle T_+ ) loses heat ( δ Q < 0 \displaystyle \delta Q < 0 ), and the cold block (initially at T − \displaystyle T_- ) gains heat ( δ Q > 0 \displaystyle \delta Q > 0 ), thus, we have

0 = − γ ∫ T + T ∗ d T T + ∫ T − T ∗ d T T = − γ log ⁡ T ∗ T + + γ log ⁡ T − T ∗ = γ log ⁡ T + T − T ∗ 2 \begin{aligned} 0 &= -\gamma \int\limits_{T_+}^{T_*} \frac{dT}{T} + \int\limits_{T_-}^{T_*} \frac{dT}{T} \\ &= - \gamma\log \frac{T_*}{T_+} + \gamma\log\frac{T_-}{T_*} \\ &= \gamma\log\frac{T_+T_-}{T_*^2} \end{aligned}

Thus we have 1 = T + T − T ∗ 2 1=\frac{T_+T_-}{T_*^2}

or

T ∗ = T + T − T_* = \sqrt{T_+T_-}

thus our engine stops working when the temperatures of the two blocks converge at the geometric average of their initial temperatures.

2 Helpful 0 Interesting 1 Brilliant 0 Confused

I find it interesting that this problem can be solved without invoking any notion of time. So if we wanted to know what the power output of the engine was for any particular temperature differential, we would presumably have to acknowledge some other parameters or constraints that are irrelevant to this particular problem.

Steven Chase - 4 years, 6 months ago

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That's the case with a lot of statistical mechanics, calculating steady states is easier than solving the dynamics. Likewise, calculating equilibrium values is usually easier than calculating out of equilibrium values.

Josh Silverman Staff - 4 years, 6 months ago

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