A thermometer reading 7 degrees C is brought into a room with a constant temperature of 29 degrees C.

If the thermometer reads 15 degrees C after 4 minutes, what will it read in 6 minutes? 11 minutes?

So I guess I need to put into newton's law of cooling:
u(t) = T + (u sub zero - T ) e^kt

How do I solve for k so can plug in my T's. and what is u sub zero? initial what?

Still lost. I put in 6 and lost a variable.

T(6) = 29 - (29 - 7) e^-0.678
29 - 22e^-0.678
-29 = -22e^-0.678
ln 1.318181818 = e^-0.678
0.276253377 = -0.678

To solve for the constant k in Newton's Law of Cooling, you need to have two sets of data points with the temperature (u) and time (t). In this case, we have two data points: u(0) = 7 degrees C at t = 0, and u(4) = 15 degrees C at t = 4 minutes.

Using these values, we can set up two equations:

First equation: 7 = T + (u(0) - T)e^(k * 0) = T
Second equation: 15 = T + (u(0) - T)e^(k * 4)

Simplifying the second equation, we get:

15 = T + (7 - T)e^(4k)
8 = (7 - T)e^(4k)
8/(7 - T) = e^(4k)

Now, we can take the natural logarithm of both sides to solve for k:

ln(8/(7 - T)) = 4k

Once you solve for k, you can substitute the value of k into the formula to find the temperature at different time intervals.

However, u sub zero represents the initial temperature at t = 0. In this case, u sub zero would be 7 degrees C.

To solve for the constant "k" in Newton's Law of Cooling, you need to have additional data, specifically the temperature readings at two different times. In this case, we only have the initial temperature (7 degrees C) and the temperature after 4 minutes (15 degrees C).

Therefore, we cannot directly solve for "k" with the given information. However, we can approximate it by assuming that the rate of cooling is constant over the short time interval (4 minutes) and use this approximation to predict future temperature readings.

Firstly, let's take a closer look at the equation u(t) = T + (u sub zero - T) * e^(-kt):

- u(t) represents the temperature at time "t."
- T is the ambient temperature of the surrounding environment (in this case, 29 degrees Celsius).
- u sub zero indicates the initial temperature of the object (in this case, 7 degrees Celsius).
- k is the constant that affects the rate at which the object cools.

Since we know the initial temperature (7 degrees Celsius) and the ambient temperature (29 degrees Celsius), we can rewrite the equation as:

u(t) = 29 + (7 - 29) * e^(-kt)

Now, let's use the available data to approximate the value of "k" and then use that to predict the temperature at different time intervals.

Given that the temperature after 4 minutes is 15 degrees Celsius, we can substitute these values into the equation:

15 = 29 + (7 - 29) * e^(-4k)

Now, to find the value of "k," we need to isolate it. Subtracting 29 from both sides, we have:

-14 = -22 * e^(-4k)

Dividing both sides by -22, we get:

e^(-4k) = 14/22

To solve for "k," we can take the natural logarithm of both sides:

-4k = ln(14/22)

Dividing both sides by -4, we find:

k ≈ -0.156

Now that we have an approximate value for "k," we can use it to predict the temperature at 6 minutes and 11 minutes.

For t = 6 minutes:

u(6) = 29 + (7 - 29) * e^(-0.156 * 6)

For t = 11 minutes:

u(11) = 29 + (7 - 29) * e^(-0.156 * 11)

Plugging these values into a calculator, you can find the approximate temperatures at 6 minutes and 11 minutes using the given values for T (29 degrees Celsius), u sub zero (7 degrees Celsius), and the approximate value of k (-0.156).

well I put the law like this:

T(t) = Ts - (Ts-To)e^(-kt)
Ts = T surroundings = 29
To = initial Temp = 7

now put in at 4 min
T(4) = 15 = 29 - (29-7)e^-4k
15 = 29 - 22 e^(-4k)
-14 = 22 e^-4k
- .636363... = e^-4k
ln (-.636363 ... ) = -4k
-4k = -.452
k = .113
so you go on and put 6 minutes in for t