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| Both sides previous revision Previous revision | |||
| 183_notes:examples:an_electric_heater [2014/10/28 04:41] – pwirving | 183_notes:examples:an_electric_heater [2014/10/28 13:59] (current) – pwirving | ||
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| 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 | ||