Question

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 we measure temperature in degrees Celsius and time in minutes, the constant of proportionality k equals 0.4. Suppose the ambient temperature TA(t) is equal to a constant 38 degrees Celsius. Write the differential equation that describes the time evolution of the temperature T of the object. (a) dT dt = $ Correct: Your answer is correct. 0.4(38-T) Suppose the ambient temp TA(t) = 38cos( π 30 t) degrees Celsius (time measured in minutes). Write the DE that describes the time evolution of temperature T of the object. (b) dT dt = Correct: Your answer is correct. If we measure time in hours the differential equation in part (b) changes. What is the new differential equation? Use the letter s to denote time in hours. (c) dT ds = If we measure time in hours and we also measure temperature in degrees Fahrenheit, the differential equation in part (c) changes even more. What is the new differential equation? Use the letter F to denote temperature in degrees Fahrenheit.

Answer 1

Answer:

Please see attachment

Explanation:

Please see attachment

Answer 2

In this exercise we have to use the knowledge of thermodynamics to calculate the functions that best correspond to temperature, in this way we find that

a)$\frac{dT}{dt}=-(0.4)(T-66)$

b) $\frac{dT}{dt}=-(0.4)(T-(66)cos(\pi t/30))$

c) $\frac{dT}{dt}=(24)(T-(66)cos(\pi t /1800))$

d) $T_A=(24)(T-(150.8)cos(\pi t /1800)$

a) According the Newton's law of cooling the rate of the heat loss from a body is directly proportional to the temperature difference between the body and the surounnding. Therefore,

$\frac{dT}{dt} = k(T-T_A)$

Here, K is the proportionality constant. Thus,:

$\frac{dT}{dt}=k(T-T_A)\\=-(0.4)(T-66)$

b) The ambient temperature is:

$T_A=(66)cos(\pi t/30)$

Substitute, the value of the ambient temperature in the equation:

$\frac{dT}{dt}=-(0.4)(T-(66)cos(\pi t/30))$

c) For the time in hours,

$K=24 hr^{-1$

Therefore,

$\frac{dT}{dt}=-(24)(T-(66)cos(\pi/1800)\\=(24)(T-(66)cos(\pi t /1800))$

d) The conversion formula for degree Celsius and the fahrenheit is:

$F=9/5C+32$

Therefore,

$T_A=(66(9/5)+32)cos(\pi t/30)\\=(150.8)cos(\pi t /30)\\=(24)(T-(150.8)cos(\pi t /1800)$

See more about Temperature at brainly.com/question/15267055