Complete Question
The complete question is shown on the first uploaded image
Answer:
The time taken is [tex]t = 6.89 \ hrs[/tex]
Explanation:
From the question we are told that
The value of [tex]k = 0.15hr^{-1}[/tex]
The the interior temperature is [tex]u(t = 0) = 70 ^oF[/tex]
The external temperature is [tex]T = 11 ^oF[/tex]
The required interior temperature is [tex]u_{t} = 32 ^oF[/tex]
The newton cooling law is
[tex]\frac{du}{dt} = -k (u -T)[/tex]
=> [tex]\frac{du}{u-T} = - kdt[/tex]
Now integrate both sides we have
[tex]\int\limits \frac{du}{u-T} =\int\limits - kdt[/tex]
[tex]ln (u - T) = -kt + c[/tex]
Since c is a constant lnC = c will also give a constant so
[tex]ln (u - T) = -kt + ln C[/tex]
=> [tex]\frac{(u -T)}{C} = e^{-kt}[/tex]
=> [tex]u = T + Ce^{-kt}[/tex]
substituting value
[tex]70 = 11 +Ce^{-0* 0.15}[/tex]
[tex]C = 59[/tex]
Hence
[tex]u_t = 11 + 59 e^{-0.15 *t}[/tex]
[tex]32= 11 + 59 e^{-0.15 *t}[/tex]
=> [tex]t = -\frac{ln(\frac{21}{59} )}{0.15}[/tex]
[tex]t = 6.89 \ hrs[/tex]
