# Application of first order differential equation-tomorrow test

• May 29th 2009, 08:17 AM
sanikui
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.
• May 29th 2009, 09:05 AM
skeeter
Quote:

Originally Posted by sanikui
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.

$\displaystyle \frac{dQ}{dt} = k(Q - 27)$

$\displaystyle \frac{dQ}{Q-27} = k \, dt$

$\displaystyle \ln(Q-27) = kt + C$

$\displaystyle Q = 27 + Ae^{kt}$

at $\displaystyle t = 4$, $\displaystyle Q = 67$

$\displaystyle 67 = 27 + Ae^{4k}$

$\displaystyle 40 = Ae^{4k}$

at $\displaystyle t = 9$, $\displaystyle Q = 47$

$\displaystyle 47 = 27 + Ae^{9k}$

$\displaystyle 20 = Ae^{9k}$

$\displaystyle \frac{40}{20} = \frac{Ae^{4k}}{Ae^{9k}}$

$\displaystyle 2 = e^{-5k}$

solve for k , then solve for A and finish.
• May 29th 2009, 09:09 AM
Danneedshelp
haha, too fast for me. I just finished the problem. The work above is correct...or at least the same as what I got.
• May 29th 2009, 09:19 AM
sanikui
Quote:

Originally Posted by skeeter
$\displaystyle \frac{dQ}{dt} = k(Q - 27)$

$\displaystyle \frac{dQ}{Q-27} = k \, dt$

$\displaystyle \ln(Q-27) = kt + C$

$\displaystyle Q = 27 + Ae^{kt}$

at $\displaystyle t = 4$, $\displaystyle Q = 67$

$\displaystyle 67 = 27 + Ae^{4k}$

$\displaystyle 40 = Ae^{4k}$

at $\displaystyle t = 9$, $\displaystyle Q = 47$

$\displaystyle 47 = 27 + Ae^{9k}$

$\displaystyle 20 = Ae^{9k}$

$\displaystyle \frac{40}{20} = \frac{Ae^{4k}}{Ae^{9k}}$

$\displaystyle 2 = e^{-5k}$

solve for k , then solve for A and finish.

Okay, i get it...hope tomorrow i can do this type of question.
Thanks everyone. V^^