Heat conduction through spheres (1 Viewer)

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Sir,
1) Two spheres of different materials, one with double the radius and 1/4th wall thickness of the other are filled with ice. If the time taken for complete melting of ice in the large sphere is 25 minutes and that in smaller sphere is 16 minutes, what is the ratio of the thermal conductivities of larger one to smaller one?
I solved it in the following way:
Let R1 and R2 be the radius of the larger sphere and smaller sphere respectively. Let d1 and d2 be the thickness of the larger sphere and smaller sphere respectively. Let Q1 and Q2 be the amount of heat conducted by the larger sphere and smaller sphere respectively. Let t1 and t2 be the time taken for melting of ice by the larger sphere and smaller sphere respectively. Let K1 and K2 be the thermal conductivities of the larger sphere and smaller sphere respectively.

Given that R1 = 2R2, d1 = d2/4, t1 = 25 minutes, t2 = 16 minutes
Q1 = K1(4(pi)(R1^2)t1/d1
Q2 = K2(4(pi)(R2^2)t2/d2
Assuming that same quantity of ice is melted in both the spheres we have Q1 =Q2
K1(R1^2)t1/d1 = K2(R2^2)t2/d2
By solving I get,
K1/K2 = 1/25
But the book answer is 8/25. Which is correct?

I have another doubt.
2) Two solid spheres of radii R1 and R2 are made of same material and have similar surfaces. The spheres are raised to the same temperature and the allowed to cool under identical conditions. Assuming the spheres to be perfect conductors of heat, what is the ratio of their rate of loss of heat? Also what is the ratio of their rate of cooling?
I know the answer to first question which is R1^2/R2^2, since Rate of loss of heat is proportional to surface area. I have a doubt in the 2nd question. Doesn’t rate of cooling and rate of loss of heat mean the same? In that case the answer to the 2nd question is also R1^2/R2^2. But the book answer is R2/R1.
 
Amith2006 said:
Given that R1 = 2R2, d1 = d2/4, t1 = 25 minutes, t2 = 16 minutes
Q1 = K1(4(pi)(R1^2)t1/d1
Q2 = K2(4(pi)(R2^2)t2/d2
Assuming that same quantity of ice is melted in both the spheres we have Q1 =Q2
K1(R1^2)t1/d1 = K2(R2^2)t2/d2
By solving I get,
K1/K2 = 1/25
But the book answer is 8/25. Which is correct?
Consider just one sphere (forget the problem completely) of radius R and wall thickness d. The volume [itex](4/3)\pi R^{3}[/itex] is filled with ice. For radial conduction you need to consider a slice of radius [itex]r[/itex] (where [itex]R \leq r \leq (R+d)[/itex]) and apply Fourier's heat conduction equation to it first. This gives

[tex]\frac{dQ}{dt} = k(4\pi r^2)\frac{dT}{dr}[/tex]

(the minus sign has been removed as heat is being transferred inwards)

Now [itex]dQ/dt[/itex] is same everywhere in the steady state. So you can call it a constant say [itex]P[/itex]. Now solve this equation to get P and then integrate to get Q as a function of t. Does this match the equation you have written? (t = time, T = temperature)
 
Amith2006 said:
I have another doubt.
2) Two solid spheres of radii R1 and R2 are made of same material and have similar surfaces. The spheres are raised to the same temperature and the allowed to cool under identical conditions. Assuming the spheres to be perfect conductors of heat, what is the ratio of their rate of loss of heat? Also what is the ratio of their rate of cooling?
I know the answer to first question which is R1^2/R2^2, since Rate of loss of heat is proportional to surface area. I have a doubt in the 2nd question. Doesn’t rate of cooling and rate of loss of heat mean the same? In that case the answer to the 2nd question is also R1^2/R2^2. But the book answer is R2/R1.
In the context of Newton's Law, rate of cooling means the rate of change of temperature with time, i.e. -[itex]dT/dt[/itex]. But some authors write -[itex]dQ/dt[/itex] instead. In any case, [itex]dQ = msdT[/itex] so you can get either quantity easily. Which book is this?

EDIT: Minus signs added.
 
Last edited:

Curious3141

Homework Helper
2,808
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maverick280857 said:
In the context of Newton's Law, rate of cooling means the rate of change of temperature with time, i.e. [itex]dT/dt[/itex]. But some authors write [itex]dQ/dt[/itex] instead. In any case, [itex]dQ = msdT[/itex] so you can get either quantity easily. Which book is this?
Rate of cooling is [tex]-\frac{dT}{dt}[/tex].

[tex]-\frac{dQ}{dt}[/tex] is rate of loss of heat energy.

If a textbook were mixing these up, I'd be very disappointed with the shoddy terminology.
 
427
2
Sir,
For the first question, I solved using the equation provided by you and I got the same expression of mine. So does that mean that my answer is right?
Now I am a bit confused. For the second question if I use the equation dQ/dt = ms(dT/dt), I get the answer as rate of loss of heat is proportional to R^3. But if I use the equation dQ/dt = [kA(dT/dt)]d, I get the answer as rate of loss of heat is proportional to R^2. The problem is that the first one depends upon volume and the second one depends upon area. Please help.
 
Curious3141 said:
Rate of cooling is [tex]-\frac{dT}{dt}[/tex].

[tex]-\frac{dQ}{dt}[/tex] is rate of loss of heat energy.

If a textbook were mixing these up, I'd be very disappointed with the shoddy terminology.
No mate, this isn't quite shoddy terminology. Chemical engineers and physicists and in general authors can disagree in undergraduate books (or even pre-undergrad books). Don't be disappointed :smile: I've seen some problems mention Newton's law as dQ/dt by absorbing the heat capacity in the so called convective coefficient...so its really a scaled coefficient with different units to get the heat current.

Anyway Amit, which book is this?
 
Last edited:
Amith2006 said:
Sir,
For the first question, I solved using the equation provided by you and I got the same expression of mine. So does that mean that my answer is right?
You have to tell me which book you are using first. There could be a misprint...but still..

For the second question if I use the equation dQ/dt = ms(dT/dt), I get the answer as rate of loss of heat is proportional to R^3. But if I use the equation dQ/dt = [kA(dT/dt)]d, I get the answer as rate of loss of heat is proportional to R^2. The problem is that the first one depends upon volume and the second one depends upon area. Please help.
The area ratio equals the square of the radius ratio and the volume ratio equals the cube of the radius ratio. Since you have to take the ratio, work the problem with both dQ/dt and dT/dt and see what you get. Again I have to ask you which book you're using. Look up the theory book you are using and see how they have defined "rate of cooling".

Also, you need dT/dt anyway so first compute it and compare...
 
427
2
Sir,
The book which I am using is a book I bought when I was preparing for my engineering entrance exams. Though there are some errors in it, it has taught me a lot. When knowledgeable people like you are there, these errors can very well be rectified, isn't it? Now forget everything Sir, and just tell me whether the first answer of mine is right or not?
 
Amith2006 said:
Sir,
The book which I am using is a book I bought when I was preparing for my engineering entrance exams. Though there are some errors in it, it has taught me a lot. When knowledgeable people like you are there, these errors can very well be rectified, isn't it? Now forget everything Sir, and just tell me whether the first answer of mine is right or not?
Amit, I am from India too so I think I can help you out better if you tell me the name of the book. All the other details are irrelevant. If you have done the problem as discussed here then I can't see why it should be wrong. But there is no harm in confirming it right??!!
 
427
2
Sir,
The name of the book is Krishna's Objective Physics. I don't know the author's name as the first few pages of the book are missing. Thanks for your help, Sir.
 
Amith2006 said:
Sir,
The name of the book is Krishna's Objective Physics. I don't know the author's name as the first few pages of the book are missing. Thanks for your help, Sir.
And I suppose you're in a bachelor's program right now. Please! Don't use such books. There are a large number of good theory and worked out problem books available for all levels. I would advise you to take a good look at these and also consult your instructors.
 

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