## 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
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 Quote by brunocamba 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
The given equation is,

$$\frac{d\theta}{dt} = 10(Ta − θ(t))$$

Where θ(t) presumably denotes the temperature at a given time t. Dividing both the equations by $(Ta − θ(t)$ you will have,

$$\frac{d\theta}{(Ta − θ)} = 10dt$$

This you can integrate to find the relation between time and temperature.
 Thanks! Integrating this do i get -log(Ta-θ) = 10 * t ?

## Calculus Word Problem - Thermodynamics

 Quote by brunocamba Thanks! Integrating this do i get -log(Ta-θ) = 10 * t ?
Yes!

You can also think about integrating,

$$\frac{d\theta}{\theta - Ta} = -10dt$$

Because it turns out easier to evaluate.

Recognitions:
Gold Member
 Quote by brunocamba Thanks! Integrating this do i get -log(Ta-θ) = 10 * t ?
You left out the constant of integration. You need to use the initial condition θ =100 at t = 0.
 Thanks! Yes I realized i was missing something, but I finally got it right. Thanks everyone!