Chapter 5. you must show all your work. solve the following problems. (a) (8 points) when a cold drink is taken from a refrigerator, its temperature is 5 ◦c. after 25 minutes in a 20◦c room its temperature has increased to 10◦c. (i) what is the temperature of the drink after 50 minutes? (ii) when will its temperature be 18◦c?

see explanation below

Explanation:

To do this, we need to use the Newton's law of cooling which is:

dT/dt = (k - Ts)

Where:

Ts: temperature of surroundings (In this case, 20 °C)

k: constant

t: time

Now, after we do the integrals of this, the general expression would be:

T(t) = Ce^kt + Ts  (1)

Now, with the first data, we need to calculate the value of C. This value will be the same after time has passed. You can see this as the concentration of the drink. As it's not experimenting any reaction, it's concentration remain the same, and only the temperature will change.

Now, to get the value of C, we can begin with the fact that at time = 0, temperature was 5°C so:

T(0) = Ce^k*0 + Ts

Replacing:

5 = Ce^1 + 20

5 - 20 = C*1

C = -15

Now that we have this, we can solve the first part of the problem

(i):

First, we need to get the value of k, we know the final temperature at t = 25, so we can solve for k, which is constant too, and then, calculate the temperature for t = 50 min

solving for k, with T = 10 °C, C = -15, Ts = 20 °C and t = 25 min:

10 = -15e^25k + 20

10 - 20 = -15e^25k

-10/-15 = e^25k

ln(-10/-15) = 25k

k = -0.405465/25

k = -0.0162

Now that we have k, let's calculate T after t = 50

T = -15e^(-0.0162)*50 + 20

T = -6.67 + 20

T(50) = 13.33 °C

(ii)

For this part, we only need to solve for t:

18 = -15e^(-0.0162)t + 20

18 - 20 / -15 = e^-0.0162t

0.1333 = e^-0.0162t

ln(0.1333) = -0.0162t

t = -2.0145/-0.0162

t = 124.37 min

the answer is:  

line graphs show patterns in data with variables that change over time; whereas, circle graphs show data as parts out of a whole (such as out of 100%).