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Calculus Word Problem - Thermodynamics |
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| Jun26-12, 11:51 AM | #1 |
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Calculus Word Problem - Thermodynamics
Consider an object at temperature θ in an a place which temperature is Ta. The rate of change of the temperature is given as: dθ/dt = 10(Ta − θ(t)). If the room temperature is constant at Ta=20,and the initial temperature of the object is θ(0) = 100. When will the object reach temperatures of 60,40 and 30?
Attempt: Integrating both sides of equation dθ/dt = 10(Ta − θ(t)) , but i am struggling with that. I am not sure if it is θ = 10 * Ta * t - 5 θˆ2 |
| Jun26-12, 12:05 PM | #2 |
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[tex]\frac{d\theta}{dt} = 10(Ta − θ(t))[/tex] Where θ(t) presumably denotes the temperature at a given time t. Dividing both the equations by [itex](Ta − θ(t)[/itex] you will have, [tex]\frac{d\theta}{(Ta − θ)} = 10dt[/tex] This you can integrate to find the relation between time and temperature. |
| Jun26-12, 12:17 PM | #3 |
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Thanks! Integrating this do i get -log(Ta-θ) = 10 * t ?
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| Jun26-12, 12:31 PM | #4 |
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Calculus Word Problem - Thermodynamics![]() You can also think about integrating, [tex]\frac{d\theta}{\theta - Ta} = -10dt[/tex] Because it turns out easier to evaluate. |
| Jun27-12, 08:13 PM | #5 |
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| Jun27-12, 10:35 PM | #6 |
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Thanks! Yes I realized i was missing something, but I finally got it right. Thanks everyone!
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