
#1
Apr2708, 02:52 PM

P: 277

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 icewater mixture at the other View Figure . The rod consists of 1.00m section of copper (one end in steam) joined endtoend to a length L2 of steel (one end in ice). Both sections of the rod have crosssection areas of 4.00 cm^2. The temperature of the coppersteel junction is 65.0 deg C after a steady state has been set up.
View Figure at http://session.masteringphysics.com/...4/YF1770.jpg 1. How much heat per second flows from the steam bath to the icewater 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 = 4.00×104m² · 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. 



#2
Apr2708, 03:39 PM

Admin
P: 21,625

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. 



#3
Apr2708, 03:45 PM

P: 277

How would you do that? I am not good with heat flux.
Please help me get started. 



#4
Apr2708, 03:58 PM

P: 277

Heat flow problem what did I do wrong?
How would I set up the expression?




#5
Apr2708, 04:44 PM

Admin
P: 21,625

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? 



#6
Apr2708, 04:59 PM

P: 277

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? 



#7
Apr2708, 06:24 PM

P: 277

Please! help me like a comdemned animal.




#8
Apr2808, 06:37 AM

Admin
P: 21,625

In order to have steadystate, 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.




#9
May108, 05:56 PM

P: 1

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. 



#10
May508, 06:14 AM

P: 1

How do you actually find the length?
Thanks in advance 


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