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Heat flow problem- what did I do wrong?

  • Thread starter frasifrasi
  • Start date
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A long rod, insulated to prevent heat loss along its sides, is in perfect thermal contact with boiling water (at atmospheric pressure) at one end and with an ice-water mixture at the other View Figure . The rod consists of 1.00-m section of copper (one end in steam) joined end-to-end to a length L2 of steel (one end in ice). Both sections of the rod have cross-section areas of 4.00 cm^2. The temperature of the copper-steel junction is 65.0 deg C after a steady state has been set up.

View Figure at http://session.masteringphysics.com/problemAsset/1042082/4/YF-17-70.jpg


1. How much heat per second flows from the steam bath to the ice-water mixture?

2. What is the length L2 of the steel section?

Now, here is my solution:

---------------------------------------------------------------------------------

∇q = 0
Because there is only heat flux along the rod this simplifies to:
dq/dx = 0 => q = constant
along the rod.

Heat flux is defined as
q = k·dT/dx
k is thermal conductivity

For constant q and k you find:
q = - k·ΔT/Δx

So the heat flow through the rod is
Q = A·q = -A·k·ΔT/Δx

The heat flow is constant , That means heat flow through copper section as through steel section as through the whole rod.

1.
consider copper section:
k = 400 W/Km for copper

Hence:
Q = -A·k·ΔT/Δx
= -4.00×10-4m² · 400W/Km · (65°C - 100°C) / 1m
= 5.6W

I put in 5.6, and it told me it was the wrong answer.

What did i do wrong?

Thanks.
 

Answers and Replies

Astronuc
Staff Emeritus
Science Advisor
18,543
1,685
Steel and copper have different thermal conductivities, so the temperature differentials will be different. The boiling water at 1 atm is at 100°C, which is sat temp at 1 atm, and the ice is at 0°C.

The interface is at 65°C, and the heat flux on both sides is equal, but the dT/dx is not.

Find expressions for the heat flux in the copper and steel at the interface and set them equal.
 
276
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How would you do that? I am not good with heat flux.

Please help me get started.
 
Last edited:
276
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How would I set up the expression?
 
Astronuc
Staff Emeritus
Science Advisor
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Q = A·q = -A·k·ΔT/Δx is correct.

Heat flux = q = -k ΔT/Δx. One knows the ΔT's and the length of copper.

So q (copper) = q (steel), and the k's are different.

The 5.6 seems correct, but for 1 sec the heat (energy) should be J. W is J/s, which is power.

Does the answer input require units?
 
Last edited:
276
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It requires W. I put in 5.6 and got "try again."

I don't know what to do. Shouldn't I add both q's?
 
276
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Please! help me like a comdemned animal.
 
Astronuc
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In order to have steady-state, what goes in one end must go out the other, i.e. the rate of heat transfer is constant along the length of the bars. The formula one used seems correct.
 
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Thermal Conductivity

Looks like you used a Thermal Conductivity value of 400 which is close to the value that wikipedia has. However, my book (and yours if same book p 664) uses the value of 385.0 for the thermal conductivity of copper.

...so you did it correctly. However, the answer using a thermal conductivity of 385 would be 5.39W.
 
Length of L_2

How do you actually find the length?

Thanks in advance
 
Hootenanny
Staff Emeritus
Science Advisor
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How do you actually find the length?

Thanks in advance
One would use the heat flux equation and solve for [itex]\Delta x[/itex].
 

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