# Rate of heat transfer passing from the boiling water to the ice bath

1. Dec 6, 2011

### lalahelp

1. The problem statement, all variables and given/known data
A copper rod of length 0.20 m and cross-sectional area 6.00 10-2 cm2 is connected to an iron rod with the same cross section and length 0.28 m (the figure below). One end of the copper is immersed in boiling water and the other end is at the junction with the iron. If the far end of the iron rod is in an ice bath at 0° C, find the rate of heat transfer passing from the boiling water to the ice bath. Assume there is no heat loss to the surrounding air.

What equation do I need to solve this could someone help?

2. Dec 6, 2011

### PeterO

Check this explanation

http://hop.concord.org/h1/phys/h1pm.html

which ends with a formula

OR [pasted from another site.

Basic Theory
The rate at which heat is conducted
through a material is proportional
to the area normal to the heat flow
and to the temperature gradient
along the heat flow path. For a one
dimensional, steady state heat flow
the rate is expressed by Fourier’s
equation:

Q = kA ΔT/d

Where:
k = thermal conductivity, W/m-K
Q = rate of heat flow, W
A = contact area
d = distance of heat flow
ΔT = temperature difference

The trick with the two rods is that the rate of heat flow in each must be the same, so the temperature at the Cu/Fe junction will NOT be 50 degrees.
Given that Copper conducts heat much better than Iron it will be closer to 100.
If, for example, copper conducts 9 times as well as Iron, the temp would be 90 degrees.
heat flow through copper with a 10 degree difference [100 - 90] would equal heat flow through Iron with a 90 degree difference [90 - 0].

3. Dec 7, 2011

### lalahelp

Q/ Time = (Thermal conductivity) x (Area) x (Thot - Tcold)/Thickness

thermal conductivity of copper = 401 W/m K
thermal conductivity of iron = 80 W/m K

For copper heat conduction:
Q/t = 401 * .0006 m^2 * (100C - 0C) / .20 m thickness
Q/t = 120.3 watts conduction heat loss up to iron junction

For iron conduction:
Q/t = 80 * .0006 m^2 * ( 100C - 0C) / .28 cm = 17.1 watts heat conduction from junction to ice bath.

Is my work correct??? I dont know what to do after?

4. Dec 7, 2011

### technician

The temperature difference of 100C is across both rods. You do not know the temperature at the junction of the copper and the iron.... call it ∅.
The temp difference for the copper is then (100 - ∅).... put this in your copper equation.
The temp difference across the iron is (∅-0) =∅...put this in the iron equation
You will then have 2 equations for heat flow through copper and iron. These are equal so find ∅ then substitute back to find heat flow.
I got ∅ = 87.5C and rate of heat flow 1500W

5. Dec 8, 2011

### lalahelp

How did you solve for ∅?

6. Dec 8, 2011

### technician

For the copper
dQ/dt = 401 x 0.0006^2 x (100-∅)/0.2

For the iron
dQ/dt = 80 x 0.0006^2 x (∅ -0)/0.28
These 2 equations are equal
Can you do the algebra to find ∅?
If not ask and I will do the next step

7. Dec 9, 2011

### lalahelp

Can you do the next step please