# Heat flow problem- what did I do wrong?

by frasifrasi
Tags: flow, heat
 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 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/...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.
 Admin P: 21,906 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.
 P: 277 How would you do that? I am not good with heat flux. Please help me get started.
 P: 277 Heat flow problem- what did I do wrong? How would I set up the expression?
 Admin P: 21,906 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?
 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?
 P: 277 Please! help me like a comdemned animal.
 Admin P: 21,906 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.
 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.
 P: 1 How do you actually find the length? Thanks in advance
Emeritus
One would use the heat flux equation and solve for $\Delta x$.