183_notes:examples:an_electric_heater

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183_notes:examples:an_electric_heater [2014/10/28 04:41] pwirving183_notes:examples:an_electric_heater [2014/10/28 13:59] (current) pwirving
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 ${\Delta E_{thermal}} = {\Delta E_{sys}} = 0 $ ${\Delta E_{thermal}} = {\Delta E_{sys}} = 0 $
  
-This is a steady-state situation.+This is a steady-state situation. This means that $E_{sys}$ = 0 and W = 0.
  
 ${\Delta E_{sys}} = W + Q$ + electric energy input ${\Delta E_{sys}} = W + Q$ + electric energy input
 +
 +Therefore, 0 is equal to the amount of energy that flows from the surroundings into the system, due to a temperature difference between the system and surroundings plus other energy transfers.
  
 $0 = Q + 5000 J$ $0 = Q + 5000 J$
 +
 +Solve for Q
  
 $Q = -5000 J$ $Q = -5000 J$
  
  
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  • by pwirving