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| Next revision | Previous revision | ||
| 183_notes:examples:an_electric_heater [2014/10/28 04:39] – created pwirving | 183_notes:examples:an_electric_heater [2014/10/28 13:59] (current) – pwirving | ||
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| Line 10: | Line 10: | ||
| === Lacking === | === Lacking === | ||
| - | {\Delta E_{thermal} for the heater. | + | ${\Delta E_{thermal}}$ for the heater. |
| Q, the energy transfer between the heater' | Q, the energy transfer between the heater' | ||
| Line 16: | Line 16: | ||
| === Approximations & Assumptions === | === Approximations & Assumptions === | ||
| - | + | Assume no other energy transfers. | |
| === Representations === | === Representations === | ||
| Line 25: | Line 24: | ||
| Surroundings: | Surroundings: | ||
| - | Energy Principle: ${\Delta E_{sys}} = W + Q + other energy transfers$ | + | Energy Principle: ${\Delta E_{sys}} = W + Q$ + other energy transfers |
| Line 34: | 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. |
| - | ${\Delta E_{sys}} = W + Q + electric energy input$ | + | ${\Delta E_{sys}} = W + Q$ + 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+5000J | 0=Q+5000J | ||
| + | |||
| + | Solve for Q | ||
| Q=−5000J | Q=−5000J | ||