Answer:
A) cooling constant = 0.0101365
B) [tex]\frac{df}{dt} = k ( 60 - F )[/tex]
c) F(t) = 60 + 77.46[tex]e^{0.0101365t}[/tex]
D)137.46 ⁰
Step-by-step explanation:
water temperature = 60⁰F
temperature of Bar after 20 seconds = 120⁰F
temperature of Bar after 60 seconds = 100⁰F
A) Determine the cooling constant K
The newton's law of cooling is given as
= [tex]\frac{df}{dt} = k(60 - F)[/tex]
= ∫ [tex]\frac{df}{dt}[/tex] = ∫ k(60 - F)
= ∫ [tex]\frac{df}{60 - F}[/tex] = ∫ kdt
= In (60 -F) = -kt - c
60 - F = [tex]e^{-kt-c}[/tex]
60 - F = [tex]C_{1} e^{-kt}[/tex] ( note : [tex]e^{-c}[/tex] is a constant )
after 20 seconds
[tex]C_{1}e^{-k(20)}[/tex] = 60 - 120 = -60
therefore [tex]C_{1} = \frac{-60}{e^{-20k} }[/tex] ------- equation 1
after 60 seconds
[tex]C_{1} e^{-k(60)}[/tex] = 60 - 100 = - 40
therefore [tex]C_{1} = \frac{-40}{e^{-60k} }[/tex] -------- equation 2
solve equation 1 and equation 2 simultaneously
= [tex]\frac{-60}{e^{-20k} }[/tex] = [tex]\frac{-40}{e^{-60k} }[/tex]
= 6[tex]e^{20k}[/tex] = 4[tex]e^{60k}[/tex]
= [tex]\frac{6}{4} e^{40k}[/tex] = In(6/4) = 40k
cooling constant (k) = In(6/4) / 40 = 0.40546 / 40 = 0.0101365
B) what is the differential equation satisfied
substituting the value of k into the newtons law of cooling)
60 - F = [tex]C_{1} e^{0.0101365(t)}[/tex]
F(t) = 60 - [tex]C_{1} e^{0.0101365(t)}[/tex]
The differential equation that the temperature F(t) of the bar
[tex]\frac{df}{dt} = k ( 60 - F )[/tex]
C) The formula for F(t)
t = 20 , F = 120
F(t ) = 60 - [tex]C_{1} e^{0.0101365(t)}[/tex]
120 = 60 - [tex]C_{1} e^{0.0101365(t)}[/tex]
[tex]C_{1} e^{0.0101365(20)}[/tex] = 60
[tex]C_{1} = 60 * 1.291[/tex] = 77.46
C1 = - 77.46⁰ as the temperature is decreasing
The formula for f(t)
= F(t) = 60 + 77.46[tex]e^{0.0101365t}[/tex]
D) Temperature of the bar at the moment it is submerged
F(0) = 60 + 77.46[tex]e^{0.01013659(0)}[/tex]
F(0) = 60 + 77.46(1)
= 137.46⁰
