Answer:
117.83° F
Explanation:
Using Newton's Law of Cooling which can be expressed as:
[tex]\dfrac{dT}{dt}= k(T-T_1)[/tex]
The differential equation can be computed as:
[tex]\dfrac{dT}{dt}= k(T-70)[/tex]
[tex]\dfrac{dT}{(T-70)}= kdt[/tex]
[tex]\int \dfrac{dT}{(T-70)}= \int kdt[/tex]
[tex]In|T-70| = kt +C[/tex]
[tex]T- 70 = e^{kt+C} \\ \\ T = 70+e^{kt+C} \\ \\ T = 70 + C_1e^{kt} --- (1)[/tex]
where;
[tex]C_1 = e^C[/tex]
At the initial condition, T(0)= 350
[tex]350 = 70 C_1^{k*0}[/tex]
[tex]350 -70 = C_1[/tex]
[tex]280 = C_1[/tex]
replacing [tex]C_1[/tex]= 280 into (1)
Hence, the differential equation becomes:
[tex]T(t) = 70 + 280 e^{kt}[/tex]
when;
time (t) = 1 hour
T(1) = 250
Since;
[tex]250 = 70 + 280 e^{k*1}[/tex]
[tex]180 = 280e^k \\ \\ \dfrac{180}{280}= e^k[/tex]
[tex]k = In (\dfrac{180}{280})[/tex]
k = -0.4418
Therefore;
T(t) = 70 + 280e^{(-0.4418)}t
After 4 hours, the temperature is:
T(t) = 70 + 280e^{(-0.4418)}4
T(4) = 117.83° F
