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.

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)[tex]\frac{dT}{dt}=-(0.4)(T-66)[/tex]

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

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

d) [tex]T_A=(24)(T-(150.8)cos(\pi t /1800)[/tex]

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,

[tex]\frac{dT}{dt} = k(T-T_A)[/tex]

Here, K is the proportionality constant. Thus,:

[tex]\frac{dT}{dt}=k(T-T_A)\\=-(0.4)(T-66)[/tex]

b) The ambient temperature is:

[tex]T_A=(66)cos(\pi t/30)[/tex]

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

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

c) For the time in hours,

[tex]K=24 hr^{-1[/tex]

Therefore,

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

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

[tex]F=9/5C+32[/tex]

Therefore,

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

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