For a counter-flow heat exchanger, fluids flow in opposite directions:

• At End 1 (Hot fluid inlet & Cold fluid outlet):
  \[ \Delta T_1 = T_{h, \text{in}} - T_{c, \text{out}} = 22^\circ\text{C} - 3^\circ\text{C} = 19^\circ\text{C} \]
• At End 2 (Hot fluid outlet & Cold fluid inlet):
  \[ \Delta T_2 = T_{h, \text{out}} - T_{c, \text{in}} = 6^\circ\text{C} - (-2^\circ\text{C}) = 6 + 2 = 8^\circ\text{C} \]
• The Logarithmic Mean Temperature Difference (LMTD, \(\Delta T_m\)) is defined as: \[ \Delta T_m = \frac{\Delta T_1 - \Delta T_2}{\ln\left(\frac{\Delta T_1}{\Delta T_2}\right)} = \frac{19 - 8}{\ln\left(\frac{19}{8}\right)} \]

Detailed Analysis of Transport Phenomena Laws:

• Fourier's Law of Heat Conduction (B): Heat flux is directly proportional to temperature gradient: $q = -k \frac{dT}{dx}$. (Gradient-based).
• Newton's Law of Cooling (C): Convective heat transfer is driven by the temperature difference/gradient across the thermal boundary layer: $q = h(T_s - T_\infty) = -k_f \left(\frac{\partial T}{\partial y}\right)_{y=0}$. (Gradient-based).
• Fick's Law of Mass Diffusion (D): Mass flux is directly proportional to concentration gradient: $J = -D \frac{dC}{dx}$. (Gradient-based).
• Stefan-Boltzmann Law (A): Governs thermal radiation based on absolute temperature to the fourth power: $E = \epsilon \sigma T^4$. It depends on absolute surface temperature, not a spatial gradient.

Detailed Analysis of Wall Temperature Profile:

• Official Answer Key Derivation:
  In the official examination calculation, measuring from the high-temperature surface ($110\ ^\circ\text{C}$) or taking temperature decrease across $x = 0.20\text{ m}$:
  $$T = 110 - \frac{110 - 40}{0.25} \times 0.20 = 110 - (280 \times 0.20) = 110 - 56 = 54\ ^\circ\text{C}$$
  NTA final answer key designates Option 3 ($54\ ^\circ\text{C}$) as the correct answer.
• Standard Steady-State 1D Conduction Theory Note:
  If measuring strictly $20\text{ cm}$ ($x = 0.20\text{ m}$) from the $40\ ^\circ\text{C}$ inner surface with outer at $110\ ^\circ\text{C}$:
  $$T(x) = T_1 + \frac{T_2 - T_1}{L} x = 40 + \frac{110 - 40}{0.25} (0.20) = 40 + 56 = 96\ ^\circ\text{C}$$

Detailed Analysis of Heat Exchanger Performance:

• Assertion (A) is True: A counter-flow heat exchanger delivers a much higher rate of heat transfer and thermal effectiveness than a parallel-flow exchanger for the same surface area, because the Log Mean Temperature Difference (LMTD) is significantly larger and fluid exit temperatures can cross.
• Reason (R) is True and Explains (A): In practical heat exchanger design under equivalent operating constraints, the more uniform temperature distribution in counter-flow minimizes local boundary layer degradation and enhances the effective overall heat transfer performance compared to parallel flow.