Thermodynamics: Cooling of a heated block of iron in contact with air

[walterminator]

Homework Statement

“We want to determine the time required to cool a block of iron weighing 240 grams, heated to 1500°C by forced convection in contact with air at the ambient temperature of 20°C.”
- We consider that all faces are exposed to the ambient air.
- surface = 62 cm^2
- Kconv = 40 W/(K * (m)^2)
- Cp = 440 j/(kg * K)

Relevant Equations

We have to use the formula : 𝜑S =𝜑 ∫dS and I really don’t understand why and how we use the integrals for the calculation.

Φ = 𝜑S
(dQ/dt) = k*S*dT
avec dT = (1500 + 273.15) - (40 + 273.15) = 1460 [k]
avec dQ = - 0.24 [kg] * 440 [j/(kg * k)] * 1460 [k] = - 154176 [j]
donc,
-(154176)/dt = 40 [W/(k * m²)] * 0.0062 [m²] * 1460 [k] = 362.08 [W]

 

[Chestermiller]

It goes like this: $$MC_p\frac{dT}{dt}=-kS(T-20)$$The initial temperature is 1500 C and, apparently, your final temperature is supposed to be 40C.

 

[walterminator]

It goes like this: $$MC_p\frac{dT}{dt}=-kS(T-20)$$The initial temperature is 1500 C and, apparently, your final temperature is supposed to be 40C.

Thank you but We want to determine the time required

 

[erobz]

Thank you but We want to determine the time required

You separate variables and integrate it.

 

[Chestermiller]

Thank you but We want to determine the time required

Are you saying that you don't know how to solve this linear first order ordinary differential equation and that you want. us to solve it for you?

 

[walterminator]

Are you saying that you don't know how to solve this linear first order ordinary differential equation and that you want. us to solve it for you?

Like I said : « I really don’t understand why and how we use the integrals for the calculation ». I would like explanations on how to proceed, the logic behind it, and why integrals are used. I really don’t understand. Theoretical explanations.
[Mentor Note: Insult deleted from post]

 

[walterminator]

You separate variables and integrate it.

Why we have to integrate ?

 

[walterminator]

Are you saying that you don't know how to solve this linear first order ordinary differential equation and that you want. us to solve it for you?

Oh, and I forgot to mention, I have the final solution and the details. I just don’t understand the resolution.

 

[erobz]

Why we have to integrate ?

Because the instantaneous rate of change in temperature $\frac{dT}{dt}$ depends on the temperature $T$ of the block of iron itself which is continually decreasing toward ambient temperature asymptotically.

 

[haruspex]

Why we have to integrate ?

I don’t think anyone understands what you are asking. It's a bit like, why did the chicken cross the road? The chicken crossed the road to get to the other side, and we have to integrate to find the answer.

Often, the analysis of a physical process consists of thinking about small steps and leads to an equation in which the rate of change of a variable depends on its current value: $\frac{dx}{dt}=f(x)$.
Manipulation to get all the x terms on one side and the t terms on the other gives $\frac{dx}{f(x)}=dt$.
That equation says that the time taken for x to increase by a small amount dx is $\frac{dx}{f(x)}$.
Adding up all these steps and taking the limit as the step size tends to zero is called integration: $\int\frac{dx}{f(x)}=\int dt=\Delta t$.

 

[walterminator]

I don’t think anyone understands what you are asking. It's a bit like, why did the chicken cross the road? The chicken crossed the road to get to the other side, and we have to integrate to find the answer.

Often, the analysis of a physical process consists of thinking about small steps and leads to an equation in which the rate of change of a variable depends on its current value: $\frac{dx}{dt}=f(x)$.
Manipulation to get all the x terms on one side and the t terms on the other gives $\frac{dx}{f(x)}=dt$.
That equation says that the time taken for x to increase by a small amount dx is $\frac{dx}{f(x)}$.
Adding up all these steps and taking the limit as the step size tends to zero is called integration: $\int\frac{dx}{f(x)}=\int dt=\Delta t$.

Maybe it’s simple for you, maybe it’s obvious for everyone else, but it’s a lot less so for me.
"Ignorance is the fertile soil in which knowledge grows." - Latin proverb.
Further detailed and substantiated explanations will undoubtedly make me a little less stupid.

 

[erobz]

Maybe it’s simple for you, maybe it’s obvious for everyone else, but it’s a lot less so for me.
"Ignorance is the fertile soil in which knowledge grows." - Latin proverb.
Further detailed and substantiated explanations will undoubtedly make me a little less stupid.

it’s just that we expect you to have some understanding of the Leibniz notation, and how it handles the solution to the differential equation if you are asked to do such an exercise. Why don’t you start by telling us exactly what you don’t understand? You say you don’t understand why it’s an integral, and the answer is because that’s the most straightforward way to solve the differential equation in post 2. The method of Separation of Variables.

 

[walterminator]

I don’t think anyone understands what you are asking. It's a bit like, why did the chicken cross the road? The chicken crossed the road to get to the other side, and we have to integrate to find the answer.

Often, the analysis of a physical process consists of thinking about small steps and leads to an equation in which the rate of change of a variable depends on its current value: $\frac{dx}{dt}=f(x)$.
Manipulation to get all the x terms on one side and the t terms on the other gives $\frac{dx}{f(x)}=dt$.
That equation says that the time taken for x to increase by a small amount dx is $\frac{dx}{f(x)}$.
Adding up all these steps and taking the limit as the step size tends to zero is called integration: $\int\frac{dx}{f(x)}=\int dt=\Delta t$.

Maybe it’s simple for you, maybe it’s obvious for everyone else, but it’s a lot less so for me.
"Ignorance is the fertile soil in which knowledge grows." - Latin proverb.

it’s just that we expect you to have some understanding of the Leibniz notation, and how it handles the solution to the differential equation if you are asked to do such an exercise. Why don’t you start by telling us exactly what you don’t understand? You say you don’t understand why it’s an integral, and the answer is because that’s the most straightforward way to solve the differential equation in post 2. The method of Separation of Variables.

I don’t know how it handles the solution to the differential equation and I’ve never studied differential equation.

 

[erobz]

Maybe it’s simple for you, maybe it’s obvious for everyone else, but it’s a lot less so for me.
"Ignorance is the fertile soil in which knowledge grows." - Latin proverb.


I don’t know how it handles the solution to the differential equation and I’ve never studied differential equation.

Then you have a body of knowledge to acquire. It best if you do some independent research first on the topic. We can’t be expected to catch you up that much. Look up a differential equations to figure out what they are, and the method of separation of variables. If you get stuck after doing that on a particular concept then run it by us.

 

[walterminator]

I don’t think anyone understands what you are asking. It's a bit like, why did the chicken cross the road? The chicken crossed the road to get to the other side, and we have to integrate to find the answer.

Often, the analysis of a physical process consists of thinking about small steps and leads to an equation in which the rate of change of a variable depends on its current value: $\frac{dx}{dt}=f(x)$.
Manipulation to get all the x terms on one side and the t terms on the other gives $\frac{dx}{f(x)}=dt$.
That equation says that the time taken for x to increase by a small amount dx is $\frac{dx}{f(x)}$.
Adding up all these steps and taking the limit as the step size tends to zero is called integration: $\int\frac{dx}{f(x)}=\int dt=\Delta t$.

Maybe it’s simple for you, maybe it’s obvious for everyone else, but it’s a lot less so for me.
"Ignorance is the fertile soil in which knowledge grows." - Latin proverb.

Then you have a body of knowledge to acquire. It best if you do some independent research first on the topic. We can’t be expected to catch you up that much. Look up a differential equations to figure out what they are, and the method of separation of variables. If you get stuck after doing that on a particular concept then run it by us.

Okok thank you

 

[Chestermiller]

The hard part of attacking physical is to articulate in words the physical mechanisms involved and translate these into a set of equations. Solving the equation(s) is supposed to be a gimme.

 

[haruspex]

Solving the equation(s) is supported to be a gimme.

Did you mean supposed?

 

[walterminator]

 

[jbriggs444]

View attachment 344364

This appears to be the worked solution from the book. Can you identify a particular line that is leaving you bewildered? At a guess, it is the transition between:
$$\frac {d(T_\text{iron} - T_\text{air})} {T_\text{iron}-T_\text{air}} = \frac{-KS}{m\ C_p}dt$$and$$\int_{1500}^{40}\frac{d(T_\text{iron}-T_\text{air})}{T_\text{iron}-T_\text{air}} = \frac{-KS}{m\ C_p}\int_{t_1}^{t_2} dt$$If you really wanted to make our lives easy, you could look at our $\LaTeX$ help page (https://www.physicsforums.com/help/latexhelp/) and typeset the equations yourself. As I have done for you. [Fixed one error, one style blunder and some minor style decisions while I was in there].

If this is where you are having problems understanding then try to be specific and tell us which part you do not understand.

 

[walterminator]

This appears to be the worked solution from the book. Can you identify a particular line that is leaving you bewildered? At a guess, it is the transition between:
$$\frac {d(T_\text{iron} - T_\text{air})} {T_\text{iron}-T_\text{air}} = \frac{-KS}{m\ C_p}dt$$and$$\int_{1500}^{40}\frac{d(T_\text{iron}-T_\text{air})}{T_\text{iron}-T_\text{air}} = \frac{-KS}{m\ C_p}\int_{t_1}^{t_2} dt$$If you really wanted to make our lives easy, you could look at our $\LaTeX$ help page (https://www.physicsforums.com/help/latexhelp/) and typeset the equations yourself. As I have done for you. [Fixed one error, one style blunder and some minor style decisions while I was in there].

If this is where you are having problems understanding then try to be specific and tell us which part you do not understand.

Thank you for the link but that’s my resolution and I don’t know if it’s correct or not. Here´s my code LaTex on overleaf :

 

[jbriggs444]

Thank you for the link but that’s my resolution and I don’t know if it’s correct or not.

It looked right, though I did not examine it carefully.

It is poor practice to post images of text here rather than the text itself. For instance, you could have used:

\author{sofianxyz}
\date{May 2024}

\begin{document}
$S = 2.5$\times$(...

Or you could have enclosed the whole thing in a $$ pair and checked whether it rendered.

The advantage of posting text rather than images of text or of posting rendered $\LaTeX$ rather than images of equations is that it allows us to correctly quote and comment on individual portions of your content. Or to post corrections. Or to accurately read what is written. It also allows you to snip away irrelevant context and post the relevant details.

 

[walterminator]

This appears to be the worked solution from the book. Can you identify a particular line that is leaving you bewildered? At a guess, it is the transition between:
$$\frac {d(T_\text{iron} - T_\text{air})} {T_\text{iron}-T_\text{air}} = \frac{-KS}{m\ C_p}dt$$and$$\int_{1500}^{40}\frac{d(T_\text{iron}-T_\text{air})}{T_\text{iron}-T_\text{air}} = \frac{-KS}{m\ C_p}\int_{t_1}^{t_2} dt$$If you really wanted to make our lives easy, you could look at our $\LaTeX$ help page (https://www.physicsforums.com/help/latexhelp/) and typeset the equations yourself. As I have done for you. [Fixed one error, one style blunder and some minor style decisions while I was in there].

If this is where you are having problems understanding then try to be specific and tell us which part you do not understand.

No, that’s MY resolution. That’s MY code LaTex on overleaf :

View attachment 344364

My reasoning: I found out about the method of separating variables and then I put everything that had to do with temperature on one side and everything that had to do with time on the other. I thought about it and told myself that the ambient temperature remained constant, therefore that d(Tair) = 0 and that t1 = 0 sec. since I separated my variables, that d(Tair) = o, then this gave the form 1/x dx and therefore it was necessary to limit the side relating to the temperature in relation to the temperature and that on the other side t1 = 0, the antiderivative of dt2 (using the constants of the integral) gave t2, the time necessary to cool this block of iron exposed to the ambient air by forced convection.

 

[walterminator]

(using the constants of the integral)

(by removing the constants from the integral)*

 

[haruspex]

Am I right in thinking that you evaluated this:
$$\int_{1500}^{40}\frac{d(T_\text{iron}-T_\text{air})}{T_\text{iron}-T_\text{air}} $$
by summing numerically?
The integral can be done algebraically:
$[\ln(T-T_{air})]_{T_{initial}}^{T_{final}}$

 

[walterminator]

Am I right in thinking that you evaluated this:
$$\int_{1500}^{40}\frac{d(T_\text{iron}-T_\text{air})}{T_\text{iron}-T_\text{air}} $$
by summing numerically?
The integral can be done algebraically:
$[\ln(T-T_{air}]_{T_{initial}}^{T_{final}}$

I don’t understand at all what you are saying.

 

[haruspex]

I don’t understand at all what you are saying.

It is a standard integral, $\int\frac{dx}x=\ln(|x|)$.

 

[walterminator]

It is a standard integral, $\int\frac{dx}x=\ln(|x|)$.

Yes ! That’s what I used but I’ve skipped this Line.

 

[walterminator]

It is a standard integral, $\int\frac{dx}x=\ln(|x|)$.

No, that’s MY resolution. That’s MY code LaTex on overleaf :


My reasoning: I found out about the method of separating variables and then I put everything that had to do with temperature on one side and everything that had to do with time on the other. I thought about it and told myself that the ambient temperature remained constant, therefore that d(Tair) = 0 and that t1 = 0 sec. since I separated my variables, that d(Tair) = o, then this gave the form 1/x dx and therefore it was necessary to limit the side relating to the temperature in relation to the temperature and that on the other side t1 = 0, the antiderivative of dt2 (using the constants of the integral) gave t2, the time necessary to cool this block of iron exposed to the ambient air by forced convection.

(using the constants of the integral)

then this gave the form 1/x dx and therefore it was necessary to limit the side relating to the temperature in relation to the temperature

 

[haruspex]

ok, I see it is tucked down the end of that near-unreadable LaTeX.
So is there anything you still need?
I think I got a slightly bigger answer, around 1800 seconds.

 

[walterminator]

ok, I see it is tucked down the end of that near-unreadable LaTeX.
So is there anything you still need?
I think I got a slightly bigger answer, around 1800 seconds.

Non merci, notant votre attitude hautaine, presque humiliante (near-unreadable LaTex wow), à la frontière du déshonorable et de votre rabaissement éclatant de mon anhanement accompli pour la résolution du problème, je préfère vous laissez entre vous et votre turpitude et votre manque de retenue.
En effet, il est flagrant que nous ne partageons pas les mêmes valeurs de bieséance, de courtoise, de correction et du minimum de politesse et d’apprentissage que nous sommes censés dispenser et diffuser sans aucune mesure lors de discussions de niveau académique et universitaire.
Dans ce cas l’obséquiosité ne serait pas de trop mais force est de constater qu’elle fait pénurie sur ce forum ou à tout le moins, à mon égard.

Notez aussi que l’éducation et l’aménité sont des prérequis qui sont donc par définition élémentaires dans le cadre de la vie civique en générale, qui s’apprennent et qu’il n’est jamais trop tard pour commencer à apprendre.
Avant d’être savants, nous sommes ignorants (ce qui est mon cas en physique).
N’en dédaignez point la mémoire en la matière.
Bien que, comme le chantait Georges Brassens ; « Quand on est con, on est con.
Qu’on ait vingt ans, qu’on soit grands-père,
Quand on est con, on est con.

Surtout, n’hésitez pas à me blâmer et à me censurer.

Adieu,

Sofian.