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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 |