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| Both sides previous revision Previous revision Next revision | Previous revision | ||
| 183_notes:examples:an_electric_heater [2014/10/28 04:40] – pwirving | 183_notes:examples:an_electric_heater [2014/10/28 13:59] (current) – pwirving | ||
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| === Approximations & Assumptions === | === Approximations & Assumptions === | ||
| - | + | Assume no other energy transfers. | |
| === Representations === | === Representations === | ||
| Line 34: | Line 33: | ||
| ${\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 | ||
| $Q = -5000 J$ | $Q = -5000 J$ | ||