# Thread: Application of first order differential equation-tomorrow test

1. ## Application of first order differential equation-tomorrow test

I have a problem of this type of question.
An object thrown into a large body of water cools at a rate proportional to the difference between its temperature and the water temperature. Suppose know that the the water is at a temperature of 27 degrees Celsius. After 4 minutes the object's temperature is 67 degrees, and after 9 minutes the object's temperature is 47 degrees Celsius. What was the temperature of the object when it was thrown into the water?

Differential equation is dQ/dt = -k(Q-Qs)

Q is the temperature of an object
Qs is the water temperature.

Not like other question, i can find K (constant) but in this question K ( constant ) i even cant find it out.

2. I think you can set up two equations to find k. And I'm assuming the water is kept at a constant temperature.

$\displaystyle \int^{67}_{Q_{0}} \frac {dQ}{Q-27} = -k \int^{4}_{0}dt$

$\displaystyle \int^{47}_{Q_{0}} \frac {dQ}{Q-27} = -k \int^{9}_{0}dt$