Differential Equations of Cooling Laws

In summary: Okay, now I'm getting confused. They are saying the rate at which heat is supplied (R) is equal to the sum of the storage rate (C) and the loss rate (h). So R+h=C? And the loss rate is just like, how much heat is lost every second? And what does this have to do with the heat capacity?...Anyway, I'm pretty sure I'm not understanding this whole thing and I would appreciate any help.
  • #1
strikemaster
3
0

Homework Statement


This is the most confusing problem I have ever seen in my life. I just plain do not know what it's saying and what exactly a proportionality constant is.

When you turn on an electric heater, such as "burner" on a stove, its temperature increases rapidly at first, then more slowly, and finally approaches a constant high temperature. As the burner warms up, heat supplied by the electricity goes to two places.
i.) Storage in the heater materials, thus warming the heater
ii.) Losses to the room
Assume that heat is being supplied at a constant rate, R. The rate at which heat is stored is directly proportional to the rate of change of temperature. Let T be the number of degrees above room temperature. Let t be time in seconds. Then the storage rate is C(dT/dt). The proportionality constant, C (calories per degree), is called the heat capacity of the heater materials. According to Newton's law of cooling, the rate at which heat is lost to the room is directly proportional to T. The (positive) proportionality constant, h, is called the heat transfer coefficient.

a.) The rate at which heat is supplied to the heater is equal to the sum of the storage rate and the loss rate. Write a differential equation that expresses this fact.
b.) Separate the variables and integrate the differential equation. Explain why the absolute value sign is not necessary in this case. Transform the answer so that temperature, T, is in terms of time, t. Use the initial condition that T=0 when t=0.
c.) Suppose that heat is supplied at a rate, R=50 cal/sec. Assume that the heat capacity is C=2 cal/deg, and that the heat transfer coefficient is h=.04 (cal/sec)/deg. Substitute these values to get T in terms of t alone.

The rest of the problem seems easy enough.

Homework Equations



We are doing differential equations, so no real equations.

The Attempt at a Solution



I am just having trouble understanding the problem and the vocabulary. It wasn't addressed earlier in the section so I am confused. The thing I am having most trouble comprehending is what exactly my variables should be. Do I have to create more than what is given, because the "rate of change of heat stored" (dT/dt) is equal to the "rate of change of temperature" (dT/dt) and they are in the same equation, aren't they? Also, what are these proportionality constants, the constants used when two variables are directly proportional? And I don't understand how the storage rate is C(dT/dt), unless this means that C=(dT/dt), not multiplied. In any case, it doesn't make sense. Would anyone like to get me started?
 
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  • #2
I'm getting dT/dt = (R+hT)/C as my differential equation. Correct?
rate of heat lost to room= -hT
R=C(dT/dt) - hT
R+hT=C(dT/dt)
...
but then when I try to do part B, I get really confused and the math is very long. I'm pretty sure I get the wrong answer. And the C's are confusing since I used to represent constant. Let's say b is the constant... if there is one.

integral (dT/(R+hT)) = integral (dt/C)
(1/h)*integral((h*dT)/(R+hT))=ln abs(c*t) + b
get rid of abs on both
ln(R+hT) = h*ln(ct) + hb hb=still a constant so let's just say b
e^(h*ln(ct))+b) = R + hT
simplify e and since e^b is constant, let's just say that constant is b
b*e^(h*ln(ct)) = R + hT
b (ct)^h = R + hT
fully simplified gives
T = b(ct)^h - R, all over h
That cannot be right, can it? especially when you plug in 0 for T and t, it gives that R = 0 as well. So I don't know where did I go wrong.. probably at the beginning. Any help appreciated.
 
  • #3
strikemaster said:

Homework Statement


This is the most confusing problem I have ever seen in my life. I just plain do not know what it's saying and what exactly a proportionality constant is.

When you turn on an electric heater, such as "burner" on a stove, its temperature increases rapidly at first, then more slowly, and finally approaches a constant high temperature. As the burner warms up, heat supplied by the electricity goes to two places.
i.) Storage in the heater materials, thus warming the heater
ii.) Losses to the room
Assume that heat is being supplied at a constant rate, R.
So the hear supplied up to time t is Rt.

The rate at which heat is stored is directly proportional to the rate of change of temperature. Let T be the number of degrees above room temperature. Let t be time in seconds. Then the storage rate is C(dT/dt). The proportionality constant, C (calories per degree), is called the heat capacity of the heater materials.
Okay, "directly proportional to the rate of change of temperature", in symbols, is just C(dT/dt) as they say.

According to Newton's law of cooling, the rate at which heat is lost to the room is directly proportional to T. The (positive) proportionality constant, h, is called the heat transfer coefficient.
So the rate at which heat is lost to the room is dH/dt= h T where "H" is the amount of heat.

a.) The rate at which heat is supplied to the heater is equal to the sum of the storage rate and the loss rate. Write a differential equation that expresses this fact.
That sounds reasonable! Since we were told earlier that the "rate at which heat is supplied" is a constant, r, we have C dT/dt+ hT= R where C, h, and R are given constants.
That can be written as C dT/dt= R- hT.

b.) Separate the variables and integrate the differential equation. Explain why the absolute value sign is not necessary in this case. Transform the answer so that temperature, T, is in terms of time, t. Use the initial condition that T=0 when t=0.
c.) Suppose that heat is supplied at a rate, R=50 cal/sec. Assume that the heat capacity is C=2 cal/deg, and that the heat transfer coefficient is h=.04 (cal/sec)/deg. Substitute these values to get T in terms of t alone.

The rest of the problem seems easy enough.

Homework Equations



We are doing differential equations, so no real equations.
? Differential equations are "real equations"!

The Attempt at a Solution



I am just having trouble understanding the problem and the vocabulary. It wasn't addressed earlier in the section so I am confused. The thing I am having most trouble comprehending is what exactly my variables should be. Do I have to create more than what is given, because the "rate of change of heat stored" (dT/dt) is equal to the "rate of change of temperature" (dT/dt) and they are in the same equation, aren't they? Also, what are these proportionality constants, the constants used when two variables are directly proportional? And I don't understand how the storage rate is C(dT/dt), unless this means that C=(dT/dt), not multiplied. In any case, it doesn't make sense. Would anyone like to get me started?
 
  • #4
Thanks. That helps until I get to the end of part B, again. For the equation, I get:
T= R(e^(-ht/c)-1)/-h
Seems reasonable.
Then, when I plug in the values, I get the equation to be -1250(e^(-.02t)-1)
When I graph, it seems alright, but then the problem asks me to predit T at times 10, 20, 50, 100, 200 after heater was turned on and then find T as t approaches infinity. This is where I get the limit to be 1250, which I would have guessed in the beginning, but that sounds extremely unreasonable. 99% steady state temperature of 1250degC above room temperature in 20 minutes/1237 seconds? Impossible. Where did I go wrong?
 

1. What are differential equations of cooling laws?

Differential equations of cooling laws are mathematical equations that describe the rate of heat loss or temperature change of an object as it cools down. These equations are used to model and predict the behavior of cooling processes in various systems, such as thermodynamics, engineering, and physics.

2. How are differential equations of cooling laws used?

Differential equations of cooling laws are used to determine the rate at which an object's temperature changes over time as it cools down. This information can then be used to design and optimize cooling systems, predict how long it will take for an object to cool down to a certain temperature, and understand the behavior of various cooling processes.

3. What are some examples of cooling laws?

There are several types of cooling laws, including Newton's Law of Cooling, Stefan-Boltzmann Law, and Fourier's Law. These laws describe different aspects of cooling processes, such as the relationship between an object's temperature and its surroundings, the rate of heat transfer through radiation, and the rate of heat conduction through a material.

4. How are differential equations of cooling laws derived?

Differential equations of cooling laws are derived using principles from calculus and thermodynamics. They involve solving for the rate of change of temperature over time, taking into account factors such as the object's initial temperature, its surroundings, and the rate of heat transfer.

5. What are the limitations of cooling laws?

Cooling laws are based on certain assumptions and may not accurately predict the behavior of cooling processes in all situations. Factors such as external disturbances, non-uniform cooling, and changes in the cooling medium can affect the accuracy of these equations. Additionally, cooling laws may not be applicable to more complex systems or processes involving phase changes.

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