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:

I solved this using initial conditions and calculus, so I hope that's what you are doing in math.  It's actually NOT calculus, just a concept that is taught in calculus.

The initial condition formula we need is

$y=Ce^{kt}$

Filling in our formula with the 2 conditions we are given:

$65=Ce^{10k}$   and   $85=Ce^{15k}$

With those 2 equations, we have 2 unknowns, the C (initial value) and the k (the constant). We know that the initial value (or starting temp) for both conditions is the same, so we solve for C in one equation, sub it into the other equation and solve for k.  If

$65=Ce^{10k}$ then

$\frac{65}{e^{10k}}=C$ which, by exponential rules is the same as

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

Since that value of C is the same as the value of C in the other equation, we sub it in:

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

Divide both sides by 65 and use the rules of exponents again to get

$\frac{85}{65}=e^{-10k+15k}$ which simplifies down to

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

Take the natural log of both sides to get

$ln(\frac{85}{65})=5k$

Do the log thing on your calculator to get

.2682639866 = 5k and divide both sides by 5 to find k:

k = .0536527973

Now that we have k, we sub THAT value in to one of the original equations to find C:

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

which simplifies down to

$65=Ce^{.536527973}$

Raise e to that power on your calculator to get

65 = C(1.710059171) and divide to solve for C:

C = 38.01038064

Now sub in k and C to the final problem when t = 23:

$y=38.01038064e^{(.0536527973)(23)}$ which simplifies a bit to

$y=38.01038064e^{1.234014338}$

Raise e to that power on your calculator to get

y = 38.01038064(3.434991111) and

finally, the temp at 23 minutes is

130.565