Answer:
a. k = -0.01014 s⁻¹
b.
c. ![\mathbf{y(t) = 60+ \dfrac{60 \sqrt{3}}{\sqrt{2}} \ e^{\dfrac{-In(\dfrac{3}{2})\ t}{40}}}](https://tex.z-dn.net/?f=%5Cmathbf%7By%28t%29%20%3D%2060%2B%20%5Cdfrac%7B60%20%5Csqrt%7B3%7D%7D%7B%5Csqrt%7B2%7D%7D%20%5C%20e%5E%7B%5Cdfrac%7B-In%28%5Cdfrac%7B3%7D%7B2%7D%29%5C%20%20t%7D%7B40%7D%7D%7D)
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
![\dfrac{dT}{dt} = k \Delta T](https://tex.z-dn.net/?f=%5Cdfrac%7BdT%7D%7Bdt%7D%20%20%3D%20k%20%5CDelta%20T)
![\dfrac{dT}{dt} = k(T_1-T_2)](https://tex.z-dn.net/?f=%5Cdfrac%7BdT%7D%7Bdt%7D%20%20%3D%20k%28T_1-T_2%29)
![\dfrac{dT}{dt} = k (T - 60)](https://tex.z-dn.net/?f=%5Cdfrac%7BdT%7D%7Bdt%7D%20%3D%20k%20%28T%20-%2060%29)
Taking the integral.
![\int \dfrac{dT}{T-60} = \int kdt](https://tex.z-dn.net/?f=%5Cint%20%5Cdfrac%7BdT%7D%7BT-60%7D%20%3D%20%5Cint%20kdt)
㏑ (T -60) = kt + C
T - 60 = ![e^{kt+C}](https://tex.z-dn.net/?f=e%5E%7Bkt%2BC%7D)
![T = 60+ C_1 e^{kt} ---- (1)](https://tex.z-dn.net/?f=T%20%3D%2060%2B%20C_1%20e%5E%7Bkt%7D%20%20----%20%20%281%29)
After 20 seconds, the temperature of the bar submersion is 120°F
T(20) = 120
From equation (1) ,replace t = 20s and T = 120
![120 = 60 + C_1 e^{20 \ k}](https://tex.z-dn.net/?f=120%20%3D%2060%20%2B%20C_1%20e%5E%7B20%20%5C%20k%7D)
![120 - 60 = C_1 e^{20 \ k}](https://tex.z-dn.net/?f=120%20-%2060%20%3D%20C_1%20e%5E%7B20%20%5C%20k%7D)
![60 = C_1 e^{20 \ k} --- (2)](https://tex.z-dn.net/?f=60%20%3D%20C_1%20e%5E%7B20%20%5C%20k%7D%20---%20%282%29)
After 1 min i.e 60 sec , the temperature = 100
T(60) = 100
From equation (1) ; replace t = 60 s and T = 100
![100 = 60 + c_1 e^{60 \ t}](https://tex.z-dn.net/?f=100%20%3D%2060%20%2B%20c_1%20e%5E%7B60%20%5C%20t%7D)
![100 - 60 =c_1 e^{60 \ t}](https://tex.z-dn.net/?f=100%20-%2060%20%3Dc_1%20e%5E%7B60%20%5C%20t%7D)
![40 =c_1 e^{60 \ t} --- (3)](https://tex.z-dn.net/?f=40%20%3Dc_1%20e%5E%7B60%20%5C%20t%7D%20---%20%283%29)
Dividing equation (2) by (3) , we have:
![\dfrac{60}{40} = \dfrac{C_1e^{20 \ k } }{C_1 e^{60 \ k}}](https://tex.z-dn.net/?f=%5Cdfrac%7B60%7D%7B40%7D%20%3D%20%5Cdfrac%7BC_1e%5E%7B20%20%5C%20k%20%7D%20%7D%7BC_1%20e%5E%7B60%20%5C%20k%7D%7D)
![\dfrac{3}{2} = e^{-40 \ k}](https://tex.z-dn.net/?f=%5Cdfrac%7B3%7D%7B2%7D%20%3D%20e%5E%7B-40%20%5C%20k%7D)
![-40 \ k = In (\dfrac{3}{2})](https://tex.z-dn.net/?f=-40%20%20%5C%20k%20%3D%20In%20%28%5Cdfrac%7B3%7D%7B2%7D%29)
- 40 k = 0.4054651
![k = - \dfrac{0.4054651}{ 40}](https://tex.z-dn.net/?f=k%20%3D%20-%20%5Cdfrac%7B0.4054651%7D%7B%2040%7D)
k = -0.01014 s⁻¹
b. What is the differential equation satisfied by the temperature y(t)?
Recall that :
![\dfrac{dT}{dt} = k \Delta T](https://tex.z-dn.net/?f=%5Cdfrac%7BdT%7D%7Bdt%7D%20%3D%20k%20%5CDelta%20T)
![\dfrac{dT}{dt} = \dfrac{- In (\dfrac{3}{2})}{40}(T-60)](https://tex.z-dn.net/?f=%5Cdfrac%7BdT%7D%7Bdt%7D%20%3D%20%5Cdfrac%7B-%20In%20%28%5Cdfrac%7B3%7D%7B2%7D%29%7D%7B40%7D%28T-60%29)
Since y is the temperature of the body , then :
(c) What is the formula for y(t)?
From equation (1) ;
where;
![T = 60+ C_1 e^{kt} ---- (1)](https://tex.z-dn.net/?f=T%20%3D%2060%2B%20C_1%20e%5E%7Bkt%7D%20%20----%20%20%281%29)
Let y be measured in degrees Fahrenheit
![y(t) = 60 + C_1 e^{-\dfrac{In (\dfrac{3}{2})}{40}t}](https://tex.z-dn.net/?f=y%28t%29%20%3D%2060%20%2B%20C_1%20e%5E%7B-%5Cdfrac%7BIn%20%28%5Cdfrac%7B3%7D%7B2%7D%29%7D%7B40%7Dt%7D)
From equation (2)
![C_1 = \dfrac{60}{e^{20 \times \dfrac{-In(\dfrac{3}{2})}{40}}}](https://tex.z-dn.net/?f=C_1%20%3D%20%5Cdfrac%7B60%7D%7Be%5E%7B20%20%5Ctimes%20%20%5Cdfrac%7B-In%28%5Cdfrac%7B3%7D%7B2%7D%29%7D%7B40%7D%7D%7D)
![C_1 = \dfrac{60}{e^{-\dfrac{1}{2} {In(\dfrac{3}{2})}}}](https://tex.z-dn.net/?f=C_1%20%3D%20%5Cdfrac%7B60%7D%7Be%5E%7B-%5Cdfrac%7B1%7D%7B2%7D%20%7BIn%28%5Cdfrac%7B3%7D%7B2%7D%29%7D%7D%7D)
![C_1 = \dfrac{60}{e^ {In(\dfrac{3}{2})^{-1/2}}}}](https://tex.z-dn.net/?f=C_1%20%3D%20%5Cdfrac%7B60%7D%7Be%5E%20%7BIn%28%5Cdfrac%7B3%7D%7B2%7D%29%5E%7B-1%2F2%7D%7D%7D%7D)
![C_1 = \dfrac{60}{\sqrt{\dfrac{2}{3}}}](https://tex.z-dn.net/?f=C_1%20%3D%20%5Cdfrac%7B60%7D%7B%5Csqrt%7B%5Cdfrac%7B2%7D%7B3%7D%7D%7D)
![C_1 = \dfrac{60 \times \sqrt{3}}{\sqrt{2}}}](https://tex.z-dn.net/?f=C_1%20%3D%20%5Cdfrac%7B60%20%5Ctimes%20%5Csqrt%7B3%7D%7D%7B%5Csqrt%7B2%7D%7D%7D)
![\mathbf{y(t) = 60+ \dfrac{60 \sqrt{3}}{\sqrt{2}} \ e^{\dfrac{-In(\dfrac{3}{2})\ t}{40}}}](https://tex.z-dn.net/?f=%5Cmathbf%7By%28t%29%20%3D%2060%2B%20%5Cdfrac%7B60%20%5Csqrt%7B3%7D%7D%7B%5Csqrt%7B2%7D%7D%20%5C%20e%5E%7B%5Cdfrac%7B-In%28%5Cdfrac%7B3%7D%7B2%7D%29%5C%20%20t%7D%7B40%7D%7D%7D)
(d) Determine the temperature of the bar at the moment it is submerged.
At the moment it is submerged t = 0
![\mathbf{y(0) = 60+ \dfrac{60 \sqrt{3}}{\sqrt{2}} \ e^{\dfrac{-In(\dfrac{3}{2})\ 0}{40}}}](https://tex.z-dn.net/?f=%5Cmathbf%7By%280%29%20%3D%2060%2B%20%5Cdfrac%7B60%20%5Csqrt%7B3%7D%7D%7B%5Csqrt%7B2%7D%7D%20%5C%20e%5E%7B%5Cdfrac%7B-In%28%5Cdfrac%7B3%7D%7B2%7D%29%5C%20%200%7D%7B40%7D%7D%7D)
![\mathbf{y(t) = 60+ \dfrac{60 \sqrt{3}}{\sqrt{2}} }](https://tex.z-dn.net/?f=%5Cmathbf%7By%28t%29%20%3D%2060%2B%20%5Cdfrac%7B60%20%5Csqrt%7B3%7D%7D%7B%5Csqrt%7B2%7D%7D%20%7D)
y(t) = 60 + 70.485
y(t) = 130.485°F