# Thread: need help with differential (problem based)

1. ## need help with differential (problem based)

The formula [tex] \frac{dT}{dt}=k(T-Tm)[/math] where k is a constant of proportionality also holds when an object absorbs heat from the surrounding medium. If a small metal bar whose initial temperature is 20oC is dropped into a container of boiling water, how long will it take the bar to reach 90oC if it is known that its temperature increased 20oC in 1 second? ? How long will it take the bar to reach 98oC ?

i'm really clueless on how to solve this problem. can anyone help? tq

2. This is Newton's Law of Cooling.

We have the initial conditions $T(0)=20, \;\ T(1)=40$

Water boils at 100 degrees C, so $T_{m}=100$. That is the temperature of the surrounding medium. In this case, the surrounding medium is boiling water.

$\frac{dT}{dt}=k(T-100)$

Separate variables:

$\frac{dT}{T-100}=kdt$

Integrate:

$ln(T-100)=kt+c$

$T-100=e^{kt+c}$

$T-100=Ce^{kt}$

Now, use the initial conditions to find k and C.

3. Why do you use Tm=100 instead of T=100?

4. Because $T_{m}$ is the temp. of the boiling water and T is the temperature at any time t.

5. i've tried solving but ended up getting ridiculous value for the answer. can you show me the steps? tq

6. The worst part is over. You have two equations with two unknowns.

This is where you use the initial conditions you were given.

We have:

$40=100+Ce^{k(1)}$

$20=100+Ce^{k(0)}$

Solve for k and C.

From the second equation we can clearly see that $C=-80$

Then we get, upon subbing into the first:

$40=100-80e^{k}$

$k=ln(3/4)$

So, our equation is $T=100-80e^{ln(3/4)t}$

or $T=100-80(3/4)^{t}$

To find how long it takes to reach 98C, set T=98 and solve for t.

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