Question

A hot metal bar is submerged in a large reservoir of water whose temperature is 60°F. The temperature of the bar 20 s after submersion is 120°F. After 1 min submerged, the temperature has cooled to 100°F. A) Determine the cooling constant k.B) What is the differential equation satisfied by the temperature F(t) of the bar?C) What is the formula for F(t)?D) Determine the temperature of the bar at the moment it is submerged.

Answer 1

Answer:

A) cooling constant =  0.0101365

B) $\frac{df}{dt} = k ( 60 - F )$

c)  F(t) = 60 + 77.46$e^{0.0101365t}$

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

= $\frac{df}{dt} = k(60 - F)$

= ∫ $\frac{df}{dt}$ = ∫ k(60 - F)

= ∫ $\frac{df}{60 - F}$ = ∫ kdt

= In (60 -F) = -kt - c

       60 - F = $e^{-kt-c}$

      60 - F = $C_{1} e^{-kt}$       ( note : $e^{-c}$ is a constant )

after 20 seconds

$C_{1}e^{-k(20)}$ = 60 - 120 = -60  

therefore $C_{1} = \frac{-60}{e^{-20k} }$ ------- equation 1

after 60 seconds

$C_{1} e^{-k(60)}$ = 60 - 100 = - 40  

therefore $C_{1} = \frac{-40}{e^{-60k} }$ -------- equation 2

solve equation 1 and equation 2 simultaneously

= $\frac{-60}{e^{-20k} }$ = $\frac{-40}{e^{-60k} }$

= 6$e^{20k}$ = 4$e^{60k}$

= $\frac{6}{4} e^{40k}$ = 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 = $C_{1} e^{0.0101365(t)}$  

F(t) = 60 - $C_{1} e^{0.0101365(t)}$

The differential equation that the temperature F(t) of the bar

$\frac{df}{dt} = k ( 60 - F )$

C) The formula for F(t)

t = 20 , F = 120

F(t ) = 60 - $C_{1} e^{0.0101365(t)}$

120 = 60 - $C_{1} e^{0.0101365(t)}$

$C_{1} e^{0.0101365(20)}$ = 60

$C_{1} = 60 * 1.291$ = 77.46

C1 = - 77.46⁰ as the temperature is decreasing

The formula for f(t)

= F(t) = 60 + 77.46$e^{0.0101365t}$

D) Temperature of the bar at the moment it is submerged

F(0) = 60 + 77.46$e^{0.01013659(0)}$

F(0) = 60 + 77.46(1)

     = 137.46⁰