Newton's Law of Cooling with no Ambient Temperature

In summary, the conversation revolves around solving a 4 part ordinary differential equation problem to determine the temperature of a dead body found in a field at different times. The equation dT/dt=k(T-Ta) is used to represent the change in temperature over time, where Ta is the ambient temperature. The solution of the differential equation is T(t)=A+50e^{-0.6}e^{-0.01t}, and by using the given temperatures at different times, A is found to be 48.6. To determine the outside temperature in the field (part C), it can be assumed that the limit of the temperature function as t approaches infinity (48.6) is the ambient temperature. Therefore, the outside temperature in
  • #1
TobyKenz1
3
0
This is a 4 part ordinary differential equation problem.

a.) You are a member of a CSI team that has discovered a dead body in a field. Upon discovery of the body you measure its temperature and find that it is 76.04058181[tex]\circ[/tex]F. 10 minutes later the temperature of the body is 73.42926519[tex]\circ[/tex]F. 10 minutes after that the temperature of the body is 71.06644821[tex]\circ[/tex]F. SET UP BUT DO NOT SOLVE the differential equation(s) and condition(s) to determine the temperature, T(t), of the body at any time t.

b.) The solution of your differential equation is T(t)=A+50e[tex]^{-0.6}[/tex]e[tex]^{-0.01t[/tex], where A is a constant. Find A.

c.) What is the outside temperature in the field?

d.)When did the individual expire (normal body temperature is 98.6[tex]\circ[/tex]F)?

For the first part, I determined all the conditions that were give, and I wrote the equation

dT/dt=k(T-Ta)

where Ta is the ambient temperature. For the second part, I got A=27.441 by using the original temperature, 76.04058181 at time equals 0.

I just don't know how to go about the third and fourth parts. Any help would be very much appreciated.
 
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  • #2
Correction: I got A=48.6, using the original temperature at time 0.
 
  • #3
Correction2: Would it hold to reason for c that since the limit of the temperature function is 48.6 as t approaches infinity that 48.6 must be the ambient temperature? Because I think I was massively overthinking this problem.

This would also make part D much easier...
 

Related to Newton's Law of Cooling with no Ambient Temperature

1. What is Newton's Law of Cooling with no Ambient Temperature?

Newton's Law of Cooling with no Ambient Temperature is a scientific principle that explains the rate at which an object loses heat when there is no external temperature affecting it. It states that the rate of temperature change of an object is directly proportional to the temperature difference between the object and its surrounding environment.

2. How does Newton's Law of Cooling with no Ambient Temperature work?

This law works by taking into account the temperature difference between the object and its surrounding environment. The larger the difference in temperature, the faster the object will lose heat. It also considers the type of material and its thermal properties, such as conductivity and specific heat, to determine the rate of cooling.

3. What are the assumptions made in Newton's Law of Cooling with no Ambient Temperature?

The main assumptions made in this law are that the object is in a perfectly insulated environment, there is no heat source or sink affecting the object, and the temperature difference between the object and its surroundings remains constant throughout the cooling process. These assumptions allow for a simplified calculation of the rate of cooling.

4. How is Newton's Law of Cooling with no Ambient Temperature used in real-life situations?

This law is commonly used in various fields such as meteorology, engineering, and food science. It can be used to predict the cooling time of hot liquids, determine the ideal storage temperature for perishable goods, and analyze the cooling rates of different materials for industrial processes.

5. What are the limitations of Newton's Law of Cooling with no Ambient Temperature?

One of the main limitations of this law is that it only applies to objects that are losing heat through convection or radiation, and not through other methods such as conduction. It also assumes a constant temperature difference, which may not always be the case in real-life situations. Additionally, this law does not take into account factors such as humidity and air movement, which can affect the rate of cooling.

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