- #1
frasifrasi
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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.
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.