Question

As part of a screening process, computer chips must be operated in an oven at 145 °C. Ten minutes after starting, the temperature is 65 °C. After 15 minutes, the temperature is 85 °C.

What is the temperature after 23 minutes?

Answer 1

Answer:

Step-by-step explanation:

This is an initial condition problem using natural logs to solve.  The formula for this is

$y=Ce^{kt}$

where y is the temp after t time, e is Euler's number, C is the initial value, and k is the constant of proportionality.  We have 2 unknowns we need to solve for before we can answer the actual question about the temp after 23 minutes.  We also can come up with 2 equations to solve for these unknowns:

$65=Ce^{10k}$      and        $85=Ce^{15k}$

Since our initial value, C, is the same for both equations, we can solve for C in one of the equations and sub it into the other in order to solve for k:

If

$65=Ce^{10k}$, then

$C=\frac{65}{e^{10k}}$, which, equivalently, is

$C=65e^{-10k}$

Subbing that value into the other equation:

$85=65e^{-10k}(e^{15k)}$

Divide both sides by 65 to get

$\frac{85}{65}=e^{-10k+15k}$ (that uses the fact that we are multiplying like bases together so we add their exponents), and

$\frac{85}{65}=e^{5k}$

Now take the natural log of both sides to get

$ln(\frac{85}{65})=ln(e^{5k})$ which simplifies to

$.2682639866=5k$ so

k = .0536527973

Now we have our k value.  We can sub it into one of our equations to solve for C now:

$65=Ce^{10(.0536527973)}$ and

$65=Ce^{.536527973}$

Raise e to that power to get

65 = C(1.710059171) so

C = 38.01038064

Now we have enough info to solve for the temp after 23 minutes:

$y=38.01038064e^{.0536527973(23)}$ and

$y=38.01038064e^{1.234014338}$

Raise e to that power to get

y = 38.01038064(3.434991111) so

y = 130.565 degrees after 23 minutes