Answer:
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
<em>k = -0.0162</em>
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
