: (III) A copper rod and an aluminum rod of the same length and cross-sectional area are attached end to end (Fig. 14–18). The copper end is placed in a furnace maintained at a constant temperature of 205°C. The aluminum end is placed in an ice bath held at a constant temperature of 0.0°C. Calculate the temperature at the point where the two rods are joined.

FIGURE 14-18 Problem 44.

In the conduction heat transfer mode, the heat energy transfer can occur in any direction from warm to cool objects. This process proceeds until the objects’ temperatures are similar.

The rate of heat conduction can be calculated by using the following expression:

\(\frac{Q}{t} = \frac{{kA\Delta T}}{l}\)

Here, Q is the heat supplied, t is the time, k is the thermal conductivity, A is the area, l is the thickness, and \(\Delta T\) is the difference in the temperature.

The temperature at the copper end is \({T_{\rm{H}}} = 205{\rm{^\circ C}}\).

The temperature at the aluminum end is \({T_{\rm{C}}} = 0{\rm{^\circ C}}\).

From table 14-4:

The thermal conductivity of the copper is \({k_{{\rm{Cu}}}} = 380\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {{\rm{s}} \cdot {\rm{m}} \cdot {\rm{^\circ C}}}}} \right.\\} {{\rm{s}} \cdot {\rm{m}} \cdot {\rm{^\circ C}}}}\).

The thermal conductivity of the aluminum is \({k_{{\rm{Al}}}} = 200\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {{\rm{s}} \cdot {\rm{m}} \cdot {\rm{^\circ C}}}}} \right.\\} {{\rm{s}} \cdot {\rm{m}} \cdot {\rm{^\circ C}}}}\).

In order to maintain the same temperature at the joint, the heat flow in the copper and aluminum rods must be the same. Therefore,

\(\begin{array}{c}{\left( {\frac{Q}{t}} \right)_{{\rm{Cu}}}} = {\left( {\frac{Q}{t}} \right)_{{\rm{Al}}}}\\\frac{{{k_{{\rm{Cu}}}}A\left( {{T_{\rm{H}}} - T} \right)}}{l} = \frac{{{k_{{\rm{Al}}}}A\left( {T - {T_{\rm{C}}}} \right)}}{l}\\{k_{{\rm{Cu}}}}\left( {{T_{\rm{H}}} - T} \right) = {k_{{\rm{Al}}}}\left( {T - {T_{\rm{C}}}} \right)\\{k_{{\rm{Cu}}}}{T_{\rm{H}}} - {k_{{\rm{Cu}}}}T = {k_{{\rm{Al}}}}T - {k_{{\rm{Al}}}}{T_{\rm{C}}}\\T\left( {{k_{{\rm{Cu}}}} + {k_{{\rm{Al}}}}} \right) = {k_{{\rm{Cu}}}}{T_{\rm{H}}} + {k_{{\rm{Al}}}}{T_{\rm{C}}}\\T = \frac{{{k_{{\rm{Cu}}}}{T_{\rm{H}}} + {k_{{\rm{Al}}}}{T_{\rm{C}}}}}{{{k_{{\rm{Cu}}}} + {k_{{\rm{Al}}}}}}\end{array}\)

Substitute the values in the above equation.

\(\begin{array}{c}T = \frac{{\left( {380\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {{\rm{s}} \cdot {\rm{m}} \cdot {\rm{^\circ C}}}}} \right.\\} {{\rm{s}} \cdot {\rm{m}} \cdot {\rm{^\circ C}}}}} \right)\left( {205{\rm{^\circ C}}} \right) + \left( {200\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {{\rm{s}} \cdot {\rm{m}} \cdot {\rm{^\circ C}}}}} \right.\\} {{\rm{s}} \cdot {\rm{m}} \cdot {\rm{^\circ C}}}}} \right)\left( {0{\rm{^\circ C}}} \right)}}{{\left( {380\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {{\rm{s}} \cdot {\rm{m}} \cdot {\rm{^\circ C}}}}} \right.\\} {{\rm{s}} \cdot {\rm{m}} \cdot {\rm{^\circ C}}}}} \right) + \left( {200\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {{\rm{s}} \cdot {\rm{m}} \cdot {\rm{^\circ C}}}}} \right.\\} {{\rm{s}} \cdot {\rm{m}} \cdot {\rm{^\circ C}}}}} \right)}}\\ = 134.3{\rm{^\circ C}}\end{array}\)

Thus, the temperature at the joint is \(134.3{\rm{^\circ C}}\).