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| 183_notes:examples:thermal_equilibrium [2014/10/28 05:02] – created pwirving | 183_notes:examples:thermal_equilibrium [2014/10/28 13:54] (current) – pwirving | ||
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| Final State: Blocks have come to thermal equilibrium, | Final State: Blocks have come to thermal equilibrium, | ||
| + | 300 gram block of aluminum which is at a temperature of 500 K | ||
| - | === Lacking === | + | 650 gram block of iron which is at a temperature of 350 K |
| + | Aluminum is placed on iron | ||
| + | Both in insulated enclosure | ||
| + | |||
| + | Specific heat capacity of aluminum is approximately 1.0 J/K/gram | ||
| + | |||
| + | Specific heat capacity of iron is approximately 0.42 J/K/gram | ||
| + | |||
| + | Metal blocks reach same temperature | ||
| + | |||
| + | |||
| + | === Lacking === | ||
| + | |||
| + | Tf common temperature | ||
| === Approximations & Assumptions === | === Approximations & Assumptions === | ||
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| The total energy of the two blocks does not change, because there is no energy transfer from or to the surroundings. | The total energy of the two blocks does not change, because there is no energy transfer from or to the surroundings. | ||
| + | |||
| + | Therefore we want to solve for Tf which is the final temperature for both blocks. | ||
| m1C1(Tf−T1i)+m2C2(Tf−T2i)=0 | m1C1(Tf−T1i)+m2C2(Tf−T2i)=0 | ||
| + | |||
| + | Substitute in values for all known variables and solve for Tf. | ||
| (300g)(1.0J/K/g)(Tf−500)+(650g)(0.42J/K/g)(Tf−350)=0 | (300g)(1.0J/K/g)(Tf−500)+(650g)(0.42J/K/g)(Tf−350)=0 | ||