Respuesta :
Answer:
[tex]T(6) = 297.33[/tex]
after 6 hours the temperature of the object will be 297.33 F
Step-by-step explanation:
We can use Newton's Law of Cooling:
[tex]T(t) = T_s + Ce^{kt}[/tex]
here,
[tex]T(t)[/tex] : The temperature that of the object (that is changing)
[tex]T_s[/tex] : Temperature of the surroundings (that is not changing) = 62
[tex]t[/tex] : time (in unit hours)
[tex]C\,\text{and}\,k[/tex] : Are constants.
In our case were given two conditions:
[tex]T(0) = 1620[/tex]: at time t = 0, the temperature of the object is 1620
and
[tex]T(1) = 1184[/tex]: at time t = 1, the temperature of the object is 1184
we're going to use these conditions to find the values of C and k
using the first equation:
[tex]T(t) = T_s + Ce^{kt}[/tex]
[tex]T(0) = 62 + Ce^{k(0)}[/tex]
[tex]1620 - 62 =C[/tex]
[tex]C = 1558[/tex]
using the second equation
[tex]T(t) = T_s + Ce^{kt}[/tex]
[tex]T(1) = 62 + 1558e^{k(1)}[/tex]
[tex]\dfrac{1184-62}{1558} = e^{k}[/tex]
[tex]\dfrac{1184-62}{1558} = e^{k}[/tex]
[tex]e^k = \dfrac{561}{779}[/tex]
[tex]k = \ln{\left(\dfrac{561}{779}\right)\approx-0.3283[/tex]
Putting these values back in the original equation:
[tex]T(t) = T_s + Ce^{kt}[/tex]
[tex]T(t) = 62 + 1558e^{t\ln{(\frac{561}{779})}}[/tex]
considering the 'ln' and 'e' cancel out, this can also be written as:
[tex]T(t) = 62 + 1558\left(\dfrac{561}{779}\right)^t[/tex]
Now we need to find what the temperature of the object will be at 6 hours.
[tex]T(t) = 62 + 1558e^{(6)\ln{(\frac{561}{779})}}[/tex]
[tex]T(6) = 297.33[/tex]
after 6 hours the temperature of the object will be 297.33 F
