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

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

`(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

`(85)/(65)=e^(-10k+15k)` which simplifies down to

`(85)/(65)=e^(5k)`

Take the natural log of both sides to get

`ln((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