Respuesta :
Answer:
a) (dT/dt) = -0.3 [T - 70]
b) (dT/dt) = -0.3 {T - [66 cos ((π/30)t)]}
c) (dT/dt) = -18 {T - [66 cos (2πt)]}
with t in hours
d) (dT/dt) = -32.4 [T - 57.6 - 118.8 cos (2πt)]
with T in Fahrenheit and t in hours
Explanation:
The Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its temperature and that of its surroundings.
If the temperature of the object = T
Temperature of the surroundings = Ambient temperature = TA(t)
(dT/dt) ∝[T - TA(t)]
Introducing the constant of proportionality, k
(dT/dt) = k [T - TA(t)]
Temperature is in degree Celsius and time is in minutes.
Because the temperature of the body is decreasing, we introduce a minus sign
(dT/dt) = -k [T - TA(t)]
a) If TA(t) = 70°C, k = 0.3
(dT/dt) = -0.3 [T - 70]
b) The ambient temp TA(t) = 66 cos ((π/30)t) degrees Celsius (time measured in minutes).
(dT/dt) = -k [T - TA(t)]
(dT/dt) = -k {T - [66 cos ((π/30)t)]}
(dT/dt) = -0.3 {T - [66 cos ((π/30)t)]}
c) If we measure time in hours the differential equation in part (b) changes.
1 hour = 60 mins
If t is now expressed in hours,
t hours = (60t) mins
(dT/dt) = -k {T - [66 cos ((π/30)t)]}
dT = -k {T - [66 cos ((π/30)t)]} dt
dT = -k {T - [66 cos ((π/30)60t)]} d(60t)
(dT) = -60k {T - [66 cos ((π/30)60t)]} dt
(dT/dt) = -60k {T - [66 cos (2πt)]}
with t in hours, k = 0.3, 60k = 18
(dT/dt) = -18 {T - [66 cos (2πt)]}
d) If we measure time in hours and we also measure temperature in degrees Fahrenheit, the differential equation in part (c) changes even more.
If T is in degree Fahrenheit
T°F = (5/9)(T°F - 32) degrees Celsius
T°F = [(5T/9) - 17.78] degrees Celsius
(dT/dt) = -60k {T - [66 cos (2πt)]}
time already converted to hours.
dT = -60k {T - [66 cos (2πt)]} dt
66 cos (2πt) degrees Celsius = {(9/5) [66 cos (2πt)] + 32} degrees Fahrenheit = {[118.8 cos (2πt)] + 57.6} degrees Fahrenheit
d[(5T/9) - 17.78] = -60k {T - [118.8 cos (2πt) + 57.6]} dt
(5/9) dT = -60k [T - 57.6 - 118.8 cos (2πt)] dt
(5/9) (dT/dt) = -60k [T - 57.6 - 118.8 cos (2πt)]
(dT/dt) = -108k [T - 57.6 - 118.8 cos (2πt)]
k = 0.3, 108k = 32.4
(dT/dt) = -32.4 [T - 57.6 - 118.8 cos (2πt)]
with T in Fahrenheit and t in hours
Hope this Helps!!!
The answers are
(a) (dT/dt) = -0.3 (T - 70)
(b) (dT/dt) = -0.3 [T - {66 cos ((π/30)t)}]
(c) (dT/dt) = -18 [T - {66 cos (2πt)}]
(d) (dT/dt) = -32.4 {T - 57.6 - 118.8 cos (2πt)}
Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its temperature and that of its surroundings.
Let the temperature of the object = T
Temperature of the surroundings = Ambient temperature = TA(t)
(dT/dt) ∝ [T - TA(t)]
(dT/dt) = -k [T - TA(t)]
where k is the constant of proportionality.
the minus sign indicates the decrease in temperature or cooling
(a) If TA(t) = 70°C,
k = 0.3(given)
(dT/dt) = -0.3 (T - 70)
(b) The ambient temp TA(t) = 66 cos ((π/30)t) degrees Celsius
(dT/dt) = -k [T - TA(t)]
(dT/dt) = -k {T - [66 cos ((π/30)t)]}
(dT/dt) = -0.3 {T - [66 cos ((π/30)t)]}
(c) If we measure time in hours then
1 hour = 60 mins
If t is now expressed in hours,
t hours = (60t) mins
(dT/dt) = -k {T - [66 cos ((π/30)t)]}
dT = -k {T - [66 cos ((π/30)t)]} dt
dT = -k {T - [66 cos ((π/30)60t)]} d(60t)
(dT) = -60k {T - [66 cos ((π/30)60t)]} dt
(dT/dt) = -60k {T - [66 cos (2πt)]}
with t in hours, k = 0.3, 60k = 18
(dT/dt) = -18 {T - [66 cos (2πt)]}
(d) If we measure time in hours and we also measure temperature in degrees Fahrenheit, then
If T is in degree Fahrenheit
T°F = (5/9)(T°F - 32) °C
(dT/dt) = -60k {T - [66 cos (2πt)]}
time already in hours.
dT = -60k {T - [66 cos (2πt)]} dt
66 cos (2πt) °C = {(9/5) [66 cos (2πt)] + 32} °F = {[118.8 cos (2πt)] + 57.6} °F
d[(5T/9) - 17.78] = -60k {T - [118.8 cos (2πt) + 57.6]} dt
(5/9) dT = -60k [T - 57.6 - 118.8 cos (2πt)] dt
(5/9) (dT/dt) = -60k [T - 57.6 - 118.8 cos (2πt)]
(dT/dt) = -108k [T - 57.6 - 118.8 cos (2πt)]
k = 0.3, 108k = 32.4
(dT/dt) = -32.4 [T - 57.6 - 118.8 cos (2πt)]
learn more about Newton's Law of Cooling:
https://brainly.com/question/22818225
