Respuesta :
Answer:
a. k = -0.01014 s⁻¹
b. [tex]\mathbf{\dfrac{dy}{dt} = - \dfrac{In(\dfrac{3}{2})}{40}(y-60)}[/tex]
c. [tex]\mathbf{y(t) = 60+ \dfrac{60 \sqrt{3}}{\sqrt{2}} \ e^{\dfrac{-In(\dfrac{3}{2})\ t}{40}}}[/tex]
d. y(t) = 130.485°F
Step-by-step explanation:
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.
(Let y be measured in degrees Fahrenheit, and t be measured in seconds.)
We are to determine :
a. Determine the cooling constant k. k = s−1
By applying the new law of cooling
[tex]\dfrac{dT}{dt} = k \Delta T[/tex]
[tex]\dfrac{dT}{dt} = k(T_1-T_2)[/tex]
[tex]\dfrac{dT}{dt} = k (T - 60)[/tex]
Taking the integral.
[tex]\int \dfrac{dT}{T-60} = \int kdt[/tex]
㏑ (T -60) = kt + C
T - 60 = [tex]e^{kt+C}[/tex]
[tex]T = 60+ C_1 e^{kt} ---- (1)[/tex]
After 20 seconds, the temperature of the bar submersion is 120°F
T(20) = 120
From equation (1) ,replace t = 20s and T = 120
[tex]120 = 60 + C_1 e^{20 \ k}[/tex]
[tex]120 - 60 = C_1 e^{20 \ k}[/tex]
[tex]60 = C_1 e^{20 \ k} --- (2)[/tex]
After 1 min i.e 60 sec , the temperature = 100
T(60) = 100
From equation (1) ; replace t = 60 s and T = 100
[tex]100 = 60 + c_1 e^{60 \ t}[/tex]
[tex]100 - 60 =c_1 e^{60 \ t}[/tex]
[tex]40 =c_1 e^{60 \ t} --- (3)[/tex]
Dividing equation (2) by (3) , we have:
[tex]\dfrac{60}{40} = \dfrac{C_1e^{20 \ k } }{C_1 e^{60 \ k}}[/tex]
[tex]\dfrac{3}{2} = e^{-40 \ k}[/tex]
[tex]-40 \ k = In (\dfrac{3}{2})[/tex]
- 40 k = 0.4054651
[tex]k = - \dfrac{0.4054651}{ 40}[/tex]
k = -0.01014 s⁻¹
b. What is the differential equation satisfied by the temperature y(t)?
Recall that :
[tex]\dfrac{dT}{dt} = k \Delta T[/tex]
[tex]\dfrac{dT}{dt} = \dfrac{- In (\dfrac{3}{2})}{40}(T-60)[/tex]
Since y is the temperature of the body , then :
[tex]\mathbf{\dfrac{dy}{dt} = - \dfrac{In(\dfrac{3}{2})}{40}(y-60)}[/tex]
(c) What is the formula for y(t)?
From equation (1) ;
where;
[tex]T = 60+ C_1 e^{kt} ---- (1)[/tex]
Let y be measured in degrees Fahrenheit
[tex]y(t) = 60 + C_1 e^{-\dfrac{In (\dfrac{3}{2})}{40}t}[/tex]
From equation (2)
[tex]C_1 = \dfrac{60}{e^{20 \times \dfrac{-In(\dfrac{3}{2})}{40}}}[/tex]
[tex]C_1 = \dfrac{60}{e^{-\dfrac{1}{2} {In(\dfrac{3}{2})}}}[/tex]
[tex]C_1 = \dfrac{60}{e^ {In(\dfrac{3}{2})^{-1/2}}}}[/tex]
[tex]C_1 = \dfrac{60}{\sqrt{\dfrac{2}{3}}}[/tex]
[tex]C_1 = \dfrac{60 \times \sqrt{3}}{\sqrt{2}}}[/tex]
[tex]\mathbf{y(t) = 60+ \dfrac{60 \sqrt{3}}{\sqrt{2}} \ e^{\dfrac{-In(\dfrac{3}{2})\ t}{40}}}[/tex]
(d) Determine the temperature of the bar at the moment it is submerged.
At the moment it is submerged t = 0
[tex]\mathbf{y(0) = 60+ \dfrac{60 \sqrt{3}}{\sqrt{2}} \ e^{\dfrac{-In(\dfrac{3}{2})\ 0}{40}}}[/tex]
[tex]\mathbf{y(t) = 60+ \dfrac{60 \sqrt{3}}{\sqrt{2}} }[/tex]
y(t) = 60 + 70.485
y(t) = 130.485°F
