Question

Three straight metal wires, AC, BC, and CD, having the same length, diameter, and thermal conductivity, are connected as shown in the figure. Heat flows from points ‘A’ and ‘B’ to point ‘C’ and from point ‘C’ to point ‘D’. Temperatures at points A, B, and D are 100 °C, 100 °C, and 40 °C, respectively. Assuming steady-state condition and no heat loss from the wires, the temperature at point ‘C’ is _________°C. (Answer in integer)

Hint:

In steady-state heat conduction, the heat flow through each wire must be the same. Use this condition to set up an equation that can help solve for the temperature at the junction point.

Answer 1

Given:
The temperature at point A \(T_A = 100^\circ C\),
The temperature at point B \(T_B = 100^\circ C\),
The temperature at point D \(T_D = 40^\circ C\),
The lengths, diameters, and thermal conductivities of the wires are identical.
Since the wires are connected in a steady-state condition, the heat flow through each wire must be the same due to the steady-state condition and no heat loss from the wires. We assume the heat flow is governed by Fourier’s Law of Heat Conduction, which is expressed as: \[ Q = \frac{k A (T_1 - T_2)}{L} \] Where:
\(Q\) is the heat flow,
\(k\) is the thermal conductivity,
\(A\) is the cross-sectional area of the wire,
\(T_1\) and \(T_2\) are the temperatures at the two ends of the wire,
\(L\) is the length of the wire.
Since all the wires have the same length, cross-sectional area, and thermal conductivity, we can simplify the equation for the heat flow \(Q\) through each wire as: \[ Q_{AC} = Q_{BC} = Q_{CD} \] The heat flow from A and B to C must be equal, and the heat flow from C to D must also be equal. Therefore, we have the following heat flow equations: \[ Q_{AC} = \frac{k A (T_A - T_C)}{L} \] \[ Q_{BC} = \frac{k A (T_B - T_C)}{L} \] \[ Q_{CD} = \frac{k A (T_C - T_D)}{L} \] Since \(Q_{AC} = Q_{BC} = Q_{CD}\), we can equate the heat flows: \[ \frac{k A (T_A - T_C)}{L} = \frac{k A (T_B - T_C)}{L} = \frac{k A (T_C - T_D)}{L} \] Simplifying the equations: \[ (T_A - T_C) + (T_B - T_C) = (T_C - T_D) \] Substitute the given temperatures: \[ (100 - T_C) + (100 - T_C) = (T_C - 40) \] Simplifying: \[ 200 - 2T_C = T_C - 40 \] Solving for \(T_C\): \[ 200 + 40 = 2T_C + T_C \] \[ 240 = 3T_C \] \[ T_C = \frac{240}{3} = 80^\circ C \] Thus, the temperature at point C is 80°C.