Thermodynamics: Introduction

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First Law of Thermodynamics

Important terms

Work done on system, W. Work is computed by original definition: the component of force moving through a distance. Examples in internal combustion engine.

Heat absorbed by system, Q. Examples in internal combustion engine.

First Law. Energy of system U -- the sum of work done on system W and heat absorbed by system Q -- is conserved, that is, U = W + Q is conserved.

Ideal Engine (Carnot cycle)


 
Heat QH > 0 added at constant TH. Piston moves out: WH < 0 . (1→2)
Expanding gas pushes piston as TH drops to TL. (2→3) [adiabatic process]
QL < 0 extracted at TL. Piston moves in: WL > 0. (3→4)
Contracting gas drags piston as TL rises to TH.(4→1) [adiabatic process]
Energy conserved: QH+WH=QL+WL.
Work done by system W =WL-WHQH=W+QL.

Efficiency

Reviewing steps in which we ignore work change during heating and cooling and heat changes during work in or work out. In other words, we assume energy is conserved in actual engine.

  1. Increases thermal energy QH at high temperature TH
  2. Exhausts thermal energy QL at low temperature TL
  3. Does work W given by energy conservation:
      QH = W + QL.

The efficiency is the work done (W) compared to the total energy input QH.

More examples

On the web

Efficiency= W/ QH (definition)
=(QH- QL)/QH (Thermo 1st law; conservation of energy
=1- QL/QH
=1 - TL/TH Q ∝T (coming attraction)

Otto Cycle

 
 
 

Latent Heat

When a material changes phase -- e.g., ice to water or water to steam -- extra energy is required to convert one 1 kg from one phase into another. These are called Latent Heat of Fusion or Latent Heat of Evaporation, respectively.
Specific Latent Heats of Fusion/Evaporation
Fusion Vaporization
Temp CkJ/kgTemp CkJ/kg
Water03451003358
Mercury-3911357394
Sulfur115604451406
Nitrogen-21025-196199

If added 1 kg of ice at 0 C to 10 kg of water at 10 C, what is temperature of final mixture?

Ice will melt and form 11 kg of water. The specific heat of water is 4.2 kJ/kg.
345=(10+1)*(10-T)*4.2
or T = 10 - 345/11./4.2 = 2.6 C.



To cite this page:
Thermodynamics: Introduction
<http://www.physics.ohio-state.edu/~wilkins.5/energy/Resources/Lectures/thermo1.html>
[Sunday, 17-Dec-2017 10:50:17 EST]
Edited by: wilkins@mps.ohio-state.edu on Wednesday, 26-Sep-2012 10:50:42 EDT