Question

A rod CD of thermal resistance 10.0 KW\(^{-1}\) is joined at the middle of an identical rod AB as shown in figure. The ends A, B and D are maintained at 200\(^{\circ}\)C, 100\(^{\circ}\)C and 125\(^{\circ}\)C respectively. The heat current in CD is P watt. The value of P is _________. 

 

Hint:

Treat steady-state heat conduction problems like DC circuits. Temperature is analogous to voltage, heat current is analogous to electric current, and thermal resistance is analogous to electrical resistance. The principle of conservation of energy at a junction is then just like Kirchhoff's junction rule.

Answer 1

Correct Answer: 2

Step 1: Understanding the Question:
This is a heat conduction problem involving a T-junction of three rods. We are given the temperatures at the free ends and the thermal resistances. We need to find the heat current (rate of heat flow) in one of the rods (CD).
Step 2: Key Formula or Approach:
The principle of thermal current is analogous to electric current. Heat current (\(H\)) is given by \(H = \frac{\Delta T}{R_{th}}\), where \(\Delta T\) is the temperature difference and \(R_{th}\) is the thermal resistance. At a junction in steady state, the total heat current flowing into the junction must equal the total heat current flowing out of it. This is analogous to Kirchhoff's current law. We will apply this principle to the junction C to find its temperature, \(T_C\).
Step 3: Detailed Explanation:
First, let's define the thermal resistances. - Thermal resistance of rod CD is \(R_{CD} = 10.0 \text{ K/W}\). (Note: The unit KW\(^{-1}\) should be K/W). - Rod AB is identical to CD, so its total thermal resistance is also \(R_{AB} = 10.0 \text{ K/W}\). - Since C is the middle of rod AB, the rod is split into two equal halves, AC and CB. The thermal resistance of each half will be half of the total resistance of AB. - \(R_{AC} = R_{CB} = \frac{R_{AB}}{2} = \frac{10.0}{2} = 5.0 \text{ K/W}\). Now, let \(T_C\) be the temperature of the junction C. Heat flows from higher to lower temperature. Assuming \(T_A>T_C>T_B\) and \(T_C>T_D\), heat flows from A to C, and from C to B and D. At steady state, heat current into C = heat current out of C. \[ H_{AC} = H_{CB} + H_{CD} \] Using the formula \(H = \frac{\Delta T}{R_{th}}\): \[ \frac{T_A - T_C}{R_{AC}} = \frac{T_C - T_B}{R_{CB}} + \frac{T_C - T_D}{R_{CD}} \] Substitute the given values: \(T_A = 200^\circ\text{C}\), \(T_B = 100^\circ\text{C}\), \(T_D = 125^\circ\text{C}\). \[ \frac{200 - T_C}{5} = \frac{T_C - 100}{5} + \frac{T_C - 125}{10} \] To solve for \(T_C\), multiply the entire equation by 10 (the LCM of 5 and 10): \[ 2(200 - T_C) = 2(T_C - 100) + 1(T_C - 125) \] \[ 400 - 2T_C = 2T_C - 200 + T_C - 125 \] \[ 400 - 2T_C = 3T_C - 325 \] \[ 400 + 325 = 3T_C + 2T_C \] \[ 725 = 5T_C \] \[ T_C = \frac{725}{5} = 145^\circ\text{C} \] Now we can find the heat current P in rod CD. \[ P = H_{CD} = \frac{T_C - T_D}{R_{CD}} \] \[ P = \frac{145 - 125}{10.0} = \frac{20}{10.0} = 2 \text{ W} \] Step 4: Final Answer:
The value of P is 2 watts.