183_notes:examples:an_electric_heater

Differences

This shows you the differences between two versions of the page.

Link to this comparison view

Both sides previous revision Previous revision
183_notes:examples:an_electric_heater [2014/10/28 04:41] pwirving183_notes:examples:an_electric_heater [2014/10/28 13:59] (current) pwirving
Line 33: Line 33:
 ΔEthermal=ΔEsys=0 ΔEthermal=ΔEsys=0
  
-This is a steady-state situation.+This is a steady-state situation. This means that Esys = 0 and W = 0.
  
 ΔEsys=W+Q + electric energy input ΔEsys=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+5000J 0=Q+5000J
 +
 +Solve for Q
  
 Q=5000J Q=5000J
  
  
  • 183_notes/examples/an_electric_heater.1414471283.txt.gz
  • Last modified: 2014/10/28 04:41
  • by pwirving