Respuesta :
Answer:
a) [tex]T(2) = 8.265[/tex]: at time t = 2 minutes the temperature will be 8.265 degress
b) [tex]6 = T(3.55)[/tex]: the temperature will be 6 degrees at time t = 3.55 minutes.
Step-by-step explanation:
When dealing with temperature changes, it's best to work with Newton's Law of Cooling.
[tex]T(t) = T_s + Ce^{kt}[/tex]
here:
[tex]T(t)[/tex] : the temperature in the room.
[tex]T_s[/tex] : ambient (or outdoor) temperature (that always remains constant, in our case: [tex]T_s = 5[/tex] )
[tex]C\,\text{and}\,k[/tex]: are constants
Our conditions are provided:
1) [tex]T(0) = 20[/tex]
2) [tex]T(1) = 12[/tex]
using the first condition
[tex]T(0) = 5 + Ce^{k(0)}\\20 = 5 + C(1)\\C = 15[/tex]
using the second condition:
[tex]T(1) = 5 + Ce^{k(1)}\\12 = 5 + Ce^{k}\\e^k = \dfrac{7}{C}[/tex]
we can use our calculated value of C to find k
[tex]e^k = \dfrac{7}{15}\\k = \ln{(\dfrac{7}{15})}\\k = -0.7621[/tex]
Finally we can put these constants back in the main equation:
[tex]T(t) = T_s + Ce^{kt}[/tex]
[tex]T(t) = 5 + 15e^{-0.7621t}[/tex] or [tex]T(t) = 5 + 15e^{\ln{(\frac{7}{15})t}[/tex]
a) Reading after one more minute?
so it's asking:
T(2) = ?
[tex]T(2) = 5 + 15e^{\ln{(\frac{7}{15})(2)}}\\T(2) = \dfrac{124}{15} \approx 8.267[/tex]
Hence, after one more minute the temperature of the room will be 8.267 degrees
b) When will it be 6 degrees?
T(t) = 6?
[tex]6 = 5 + 15e^{-0.7621t}\\\text{and solve for $t$}\\\\\dfrac{6-5}{15}=e^{\ln{(\frac{7}{15})}t}\\\ln{\left(\dfrac{1}{15}\right)} = \ln{\left(\dfrac{7}{15}\right)t}\\\ln{\left(\dfrac{1}{15}\right)} \div \ln{\left(\dfrac{7}{15}\right)} = t \approx 3.55\\[/tex]
Hence at t = 3.55 minutes the temperature of the room will be 6 degrees.
