A long rod, insulated to prevent heat loss along its sides, is in perfect therma

Question

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 (Fig. E17.62). The rod consists of a 1.00-m section of copper (one end in boiling water) joined end to end to a length L2 of steel (one end in the ice-water mixture). Both sections of the rod have cross-sectional areas of 4.00 cm2. The temperature of the copper-steel junction is 65.0o C after a steady state has been set up. (a) How much heat per second flows from the boiling water to the ice-water mixture? (b) What is the length L2 of the steel section?

Explanation / Answer

In purely conductive energy transport without energy sources or sinks the gradient of the heat flux vector equals zero
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·T/x

So the heat flow through the rod is
Q = A·q = -A·k·T/x

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·T/x
= -4.00×10^-4m² · 400W/Km · (65°C - 100°C) / 1m
= 5.6W

2. k = 50W/Km for unalloyed steel
Q = -A·k·T/x
=> x = -A·k·T/Q
= -4.00×10-4m² · 50W/Km · (0°C - 65°C) / 5.6W
= 0.232m

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