# Help with Rates of Change Homework

• lufbrajames
In summary, the temperature of the copper bit of a soldering iron varies with time after the power has been switched on. This is a first step to determine how long it takes for the temperature of the bit to reach the operating temperature at which it can melt the solder. We assume all heat produced goes directly to the bit and none is lost to the air. The heat traveling from the barrel to the bit is the sum of the heat loss from the bit and the heat stored in the bit (conservation of energy).
lufbrajames

## Homework Statement

We need to know the temperature of the copper bit of a soldering iron varies with time after the power has been switched on. This is a first step to determine how long it takes for the temperature of the bit to reach the operating temperature at which it can melt the solder. We assume all heat produced goes directly to the bit and none is lost to the air. i.e. the temperature of the bit, Theta = Theta (t), depends only on time.

3 laws of physics:

the rate of energy storage in the bit is the product of the mass m of copper, the specific heat c of copper and the rate of change in the bit.

the rate of loss of heat from the bit to the air has the form kA(theta - Thate a) where theta a is the temprature of the air, A is the (constant) cross section of the bit, and k is a constant

The heat traveling from the barrel to the bit is the sum of the heat loss from the bit and the heat stored in the bit (consesrvation of energy)

1) To which value do you expect the temprature of the bit to settle?

2) Sketch a graph of Theta with t (time)

3) Write down the differential equation which describes the cooling process.

4)Given that the solution tof the equation

dtheta/dt + a theta = b

where a and b are cpnstant, is

theta = b/a + Ce^-at

where c is a constant, write down the solution of your eqaution in part 3 which satisfies the initial condition theta = theta 0 at t = 0

## The Attempt at a Solution

1) the temperaturewill settle at theta a the temperature of the air.

2)i drew a graph that showed the temperature drop rapidly at first and then steadyout to nearly level at theta a

3) (this is where i get really stuck!) i got the following

dtheta/dt + k = A(theta - theta 0)

4)

theta = A(theta - theta 0) / k

Question 1 and 2 seem ok to me. For question 3 you want something like this:

$$\frac{d\theta}{dt}=kA(\theta-\theta_a)$$

For question 4 put the above equation in the same form as the one they give you and you can imply the solution. Or you could always try and solve it yourself but its a tricky one.

Thanks, I am still a bit confused about 4

from this equation:
$$\frac{d\theta}{dt}=kA(\theta-\theta_a)$$

i get

$$\theta=\frac{b}{k}+Ce^kt$$

when trying to put it into the form given in quation 4...

is this at all right?

lufbrajames said:
Thanks, I am still a bit confused about 4

from this equation:
$$\frac{d\theta}{dt}=kA(\theta-\theta_a)$$

i get

$$\theta=\frac{b}{k}+Ce^kt$$

Both integration variables need to be on one side each !
k and A are constants

$$\frac{d\theta}{dt}=kA(\theta-\theta_a)$$
$$\frac{d\theta}{\theta-\theta_a}=kAdt$$

Integrate left and right hand side to come to (make sure you take into account given initial conditions for theta and t !) :

$$ln ( \theta-\theta_a ) =kAt$$

and if ln(a) = b <-> a = exp(b)

you should be able to take it from here

marlon

Thanks for the help so far, i think i mite be getting in now i have:

$$\frac{d\theta}{dt}=kA(\theta-\theta_a)$$ $$\frac{d\theta}{\theta-\theta_a}=kAdt$$ $$ln ( \theta-\theta_a ) =kAt$$ $$( \theta-\theta_a ) = e^{kAt}$$

and becuase it must satisfy $$\theta = \theta_0$$ and t = 0 then:

$$e^{kA} = \theta-\theta_a$$Is this correct?

Last edited:
You're missing a constant of integration, and theta has disappered.

Also, jumping back to question 1 - if the powe has been turned ON, the bit won't settle to the ambient temperature, it will stop at the temperature such that the electrical energy being converted to thermal energy = energy lost to the air. Might want to check that.

$$e^{kA} = \theta-\theta_a + C$$ ??

also the first question says when the electricity supply is switched off, sory that i missed that bit out.

put the constant on the side integrated with respect to t, so when you exponentiate, you get a constant multiplier on that side Qexp(kAt)

ok i think i understand

$$Ce^{kAt} = \theta-\theta_a$$

i also have to sketch a graph of the solution and compare it to my original sketch in part 2...

would'nt they be exactly the same?

after sketching the graph, the next part says

Now we switch on the electricity supply.

1. Let W be the (constan) heat per second supplied to the solderin iron, produce a differential equation for $$\theta$$

2. what is the steady-state temperature of the bit?

3 sketch the graph of $$\theta$$ against time

4 if $$\theta =\theta_0$$ at t = 0 find the solution for theta as a function of time, you may assume that theta_0 = theta_a in order to sketch the graph of temperature with time

---------

i think i have the answer to part 1

$$\frac{d\theta}{dt} = kAW-(\theta - \theta_a)$$

does steady-state temperature mean average? or the temperature at which it stops rising?

for the 3rd part i believe the graph should climb slowly at first then shoot up rapidly?

this is what i have for part 4:

$$\frac{d\theta}{dt} = kAW-(\theta - \theta_a)$$

$$\frac{d\theta}{\theta - \theta_a} = kAW-dt$$

$$ln(\theta-\theta_a) = kAW-t$$

$$(\theta-\theta_a) = Ce^{kAW-t}$$

Last edited:
anyone? I am rly stugling with this topic

lufbrajames said:
$$( \theta-\theta_a ) = e^{kAt}$$

and becuase it must satisfy $$\theta = \theta_0$$ and t = 0 then:

$$e^{kA} = \theta-\theta_a$$

Is this correct?

For this bit you should get:

$$( \theta-\theta_a ) = Ce^{kAt}$$

then when $$\theta = \theta_0$$ at t=0 one would get,

$$C= \theta_0 - \theta_a$$

For question 1 of the second set you're on the right lines so the rate of change of temperature wrt time will be the heat input minus the heat lost.

$$\frac{d\theta}{dt} = W -kA(\theta - \theta_a)$$

For part two steady state means when an equilibrium is reached so in other words when the temperature does not change any more. Therefore you want to make $$\frac{d\theta}{dt}=0$$.

Part 3 will show the temperature rising then reaching a plateau.

For part 4 you will have to solve the equation and apply the conditions given.

Last edited:
What do you mean solve the equation? there are no figures givend except for:

$$\theta = \theta_0$$ at $$t = 0$$

Assume $$\theta_0 = \theta_a$$

lufbrajames said:
What do you mean solve the equation? there are no figures givend except for:

$$\theta = \theta_0$$ at $$t = 0$$

Assume $$\theta_0 = \theta_a$$

The second set of questions asks you to come up with a differential equation in question 1. For part four it asks you to solve the equation (constructed in part 1) to find $$\theta(t)$$ with the conditions given.

oh rite, like i did before,

How do i rearange $$\theta - \theta_a = Ce^{W-kAt}$$

to get A, i have figures for W K and $$\theta_a$$ and $$\theta$$

I'm not sure you've solved the equation properly. You should be coming out with:

$$\theta(t) = \frac{W+kA\theta_a}{kA} +C e^{-kAt}$$

Last edited:
$$\theta(t) = \frac{W+kA\theta_a}{kA} +C e^{kAt}$$

how did you get this? does this mean that earlier in my thread, i have also solved an equation wrong?

$$\frac{d\theta}{dt}=kA(\theta-\theta_a)$$

and solving this i got:

$$Ce^{kAt} = \theta-\theta_a$$

Last edited:
lufbrajames said:
$$\theta(t) = \frac{W+kA\theta_a}{kA} +C e^{kAt}$$

how did you get this? does this mean that earlier in my thread, i have also solved an equation wrong?

$$\frac{d\theta}{dt}=kA(\theta-\theta_a)$$

and solving this i got:

$$Ce^{kAt} = \theta-\theta_a$$

The equation you got previously was fine. What I did is rather than attempt to solve it normally which is tricky, is to compare it to the equation in question 4 of the first set and infer the solution. We can write the differential equation as:

$$\frac{d\theta}{dt}+kA\theta=W+kA\theta_a$$

For the equation in your first post we see that $$a= kA$$ and $$b=W+kA\theta_a$$

Oh I made a mistake in the post above. The equation should be

$$\theta(t) = \frac{W+kA\theta_a}{kA} +C e^{-kAt}$$

ok thanks for the help so far.

Does the kA on the bottom cancel with the kA on the top?

lufbrajames said:
ok thanks for the help so far.

Does the kA on the bottom cancel with the kA on the top?

no. you can use $$\frac{W+kA\theta_a}{kA} = \frac{W}{kA} + \theta_a$$

if you wish

How can i find the value for A? i have the follwing:

W = 50 watts (heat supplied to iron)
Operating temperature = 500 degrees C ($$\theta_0$$)
air temperature = 20 degrees C ($$\theta_a$$)
k = 1750 j/s/K/m^2

but i carnt seem to rearange it properly.

The value of A will be temperature dependent. The original question syas you may interpret it as being constant. I suggest if you really wish to find A that you consider the dimensions of k and solve it when the temperature is that of room temp. remember j/s = watts and you want m2.

## 1. What is a rate of change?

A rate of change is a measure of how one quantity changes in relation to another quantity. It is typically expressed as a ratio or a percentage and is used to describe the speed or direction of change over a specific time period.

## 2. How do I calculate the rate of change?

The rate of change is calculated by taking the difference between two values of a variable and dividing it by the difference in time between those values. This can be represented using the formula: (y2 - y1) / (x2 - x1), where y represents the change in the dependent variable and x represents the change in the independent variable.

## 3. What is the difference between average rate of change and instantaneous rate of change?

The average rate of change refers to the overall change in a quantity over a specific interval of time, while the instantaneous rate of change refers to the change at a specific moment in time. The average rate of change is calculated using multiple data points, while the instantaneous rate of change is calculated using only one data point at a specific time.

## 4. Can rates of change be negative?

Yes, rates of change can be negative. This indicates a decrease or decrease in the quantity being measured over time. A positive rate of change indicates an increase in the quantity over time.

## 5. How are rates of change used in real life?

Rates of change are used in many real-life applications, such as calculating the speed of a moving object, predicting population growth, or analyzing stock market trends. They are also used in fields such as physics, economics, and engineering to understand and predict changes in various systems.

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