First slide

Heat transfer

Question

Three rods made of the same material and having the same cross-section have been joined as shown in Fig. Each rod is of the same length. The left and right ends are kept at 0°C and 90°C, respectively. The temperature of the junction of the three rods will be

Moderate

A

45°C
B

60°C
C

30°C
D

20°C

Solution

Let A and I be the area of cross-section and the length of each rod. If k is the coefficient of thermal conductivity and t°C the temperature of the junction O, then the rates at which heat energy enters O from rods A and B are (Fig.)
$\begin{array}{l} Q_{A} = \frac{kA (90 - t)}{l} \\ and Q_{B} = \frac{kA (90 - t)}{l} \end{array}$
The rate at which heat energy flows in rod C is
$Q_{C} = \frac{kA (t - 0)}{l}$
In the steady state, rate at which heat energy enters
O = rate at which heat energy leaves O, i.e.
$\begin{array}{l} or Q_{A} + Q_{B} = Q_{C} \\ or \frac{kA (90 - t)}{l} + \frac{kA (90 - t)}{l} = \frac{kA (t - 0)}{l} \\ or (90 - t) + (90 - t) = t \end{array}$
$or 3 t = 180 or t = 60^{\circ} C$ . Hence the correct choice is (b).

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